New test - October 20, 2014 [531 marks]

New test - October 20, 2014 [531 marks]

1a. [3 marks]Let . Find an expression for in terms of m.

MarkschemeNote: All answers must be given in terms of m. If a candidate makes an error that means there is no m in their answer, do not awardthe final A1FT mark.

METHOD 1

valid approach involving Pythagoras (M1)

e.g. , labelled diagram

correct working (may be on diagram) (A1)

e.g. ,

A1 N2

[3 marks]

METHOD 2

valid approach involving tan identity (M1)

e.g.

correct working (A1)

e.g.

A1 N2

[3 marks]

Examiners reportWhile many candidates correctly approached the problem using Pythagoras in part (a), very few recognized that the cosine of anangle in the second quadrant is negative. Many were able to earn follow-through marks in subsequent parts of the question. Acommon algebraic error in part (a) was for candidates to write . In part (c), many candidates failed to use thedouble-angle identity. Many incorrectly assumed that because , then . In addition, some candidates didnot seem to understand what writing an expression "in terms of m" meant.

sin = m100∘ cos100∘

x + x = 1sin2 cos2

+ (cos100 = 1m2 )2 1 − m2− −−−−√

cos100 = − 1 − m2− −−−−√

tan = sincos

cos100 = sin 100tan 100

cos100 = m

tan 100

= 1 − m1 − m2− −−−−√sin = m100∘ sin = 2m200∘

1b. [1 mark]Let . Find an expression for in terms of m.sin = m100∘ tan 100∘

MarkschemeMETHOD 1

(accept ) A1 N1

[1 mark]

METHOD 2

A1 N1

[1 mark]


tan100 = − m

1−m2√m

− 1−m2√

tan100 = m

cos 100

= 1 − m1 − m2− −−−−√sin = m100∘ sin = 2m200∘

1c. [2 marks]Let . Find an expression for in terms of m.

MarkschemeMETHOD 1

valid approach involving double angle formula (M1)

e.g.

(accept ) A1 N2

Note: If candidates find , award full FT in parts (b) and (c), even though the values may not have appropriatesigns for the angles.

[2 marks]

METHOD 2

valid approach involving double angle formula (M1)

e.g. ,

A1 N2

[2 marks]


sin = m100∘ sin 200∘

sin 2θ = 2 sin θcosθ

sin 200 = −2m 1 − m2− −−−−√ 2m (− )1 − m2− −−−−√

cos100 = 1 − m2− −−−−√

sin 2θ = 2 sin θ cosθ 2m × m

tan 100

sin 200 = (= 2m cos100)2m2

tan 100

= 1 − m1 − m2− −−−−√sin = m100∘ sin = 2m200∘

f(x) = acos(b(x − c))

2a. [2 marks]

Let . The diagram below shows part of the graph of f , for .

The graph has a local maximum at P(3, 5) , a local minimum at Q(7, − 5) , and crosses the x-axis at R.

Write down the value of

(i) ;

(ii) .

Markscheme(i) (accept ) A1 N1

(ii) (accept , if ) A1 N1

Note: Accept other correct values of c, such as 11, , etc.

[2 marks]

Examiners reportPart (a) (i) was well answered in general. There were more difficulties in finding the correct value of the parameter c.

f(x) = acos(b(x − c)) 0 ≤ x ≤ 10

a

c

a = 5 −5

c = 3 c = 7 a = −5

−5

2b. [2 marks]Find the value of b .

Markschemeattempt to find period (M1)

e.g. 8 ,

(exact), , 0.785 [ ] (do not accept 45) A1 N2

[2 marks]

Examiners reportFinding the correct value of b in part (b) also proved difficult as many did not realize the period was equal to 8.

b = 2π

period

0.785398…

b = 2π

8π

40.785, 0.786

2c. [2 marks]Find the x-coordinate of R.

Markschemevalid approach (M1)

e.g. , symmetry of curve

(accept A1 N2

[2 marks]

Examiners reportMost candidates could handle part (c) without difficulties using their GDC or working with the symmetry of the curve althoughfollow through from errors in part (b) was often not awarded because candidates failed to show any working by writing down theequations they entered into their GDC.

f(x) = 0

x = 5 (5 ,0))

3a. [3 marks]

The following diagram shows a circular play area for children.

The circle has centre O and a radius of 20 m, and the points A, B, C and D lie on the circle. Angle AOB is 1.5 radians.

Find the length of the chord [AB].

MarkschemeNote: In this question, do not penalise for missing or incorrect units. They are not included in the markscheme, to avoid complexanswer lines.

METHOD 1

choosing cosine rule (must have cos in it) (M1)

e.g.

correct substitution (into rhs) A1

e.g. ,

A1 N2

[3 marks]

METHOD 2

choosing sine rule (M1)

e.g. ,

correct substitution A1

e.g.

A1 N2

[3 marks]

Examiners reportCandidates generally handled the cosine rule, sectors and arcs well, but some candidates incorrectly treated triangle AOB as a right-angled triangle. A surprising number of candidates changed all angles to degrees and worked with those, often leading to errors inaccuracy.

= + − 2ab cosCc2 a2 b2

+ − 2(20)(20)cos1.5202 202 AB = 800 − 800 cos1.5− −−−−−−−−−−−−√

AB = 27.26555…

AB = 27.3 [27.2, 27.3]

=sin A

a

sin B

b=AB

sin O

AOsin B

=ABsin 1.5

20sin(0.5(π−1.5))

AB = 27.26555…

AB = 27.3 [27.2, 27.3]

3b. [2 marks]Find the area of triangle AOB.

Markschemecorrect substitution into area formula A1

e.g. ,

(accept , from using 27.3)

A1 N1

[2 marks]

Examiners reportCandidates generally handled the cosine rule, sectors and arcs well, but some candidates incorrectly treated triangle AOB as a right-angled triangle. A surprising number of candidates changed all angles to degrees and worked with those, often leading to errors inaccuracy.

(20)(20)sin 1.512

(20)(27.2655504…)sin(0.5(π − 1.5))12

area = 199.498997… 199.75106 = 200

area = 199 [199, 200]

3c. [3 marks]Angle BOC is 2.4 radians.

Find the length of arc ADC.

Markschemeappropriate method to find angle AOC (M1)

e.g.

correct substitution into arc length formula (A1)

e.g. ,

(i.e. do not accept ) A1 N2

Notes: Candidates may misread the question and use . If working shown, award M0 then A0MRA1 for the answer 48.Do not then penalize in part (d) which, if used, leads to the answer

However, if they use the prematurely rounded value of 2.4 for , penalise 1 mark for premature rounding for the answer 48 in(c). Do not then penalize for this in (d).

[3 marks]

Examiners reportIn part (c), some candidates misread the question and used 2.4 as the size of angle AOC while others rounded prematurely leading tothe inaccurate answer of 48. In either case, marks were lost.

2π − 1.5 − 2.4

(2π − 3.9) × 20 2.3831853… × 20

arc length = 47.6637…

arc length = 47.7 (47.6, 47.7] 47.6

A C = 2.4OA CO 679.498…

A CO

3d. [3 marks]Angle BOC is 2.4 radians.

Find the area of the shaded region.

Markschemecalculating sector area using their angle AOC (A1)

e.g. , ,

shaded area = their area of triangle AOB + their area of sector (M1)

e.g. ,

(accept from using 199)

A1 N2

[3 marks]

Examiners reportPart (d) proved to be straightforward and candidates were able to obtain full FT marks from errors made in previous parts.

(2.38…)( )12

202 200(2.38…) 476.6370614…

199.4989973… + 476.6370614… 199 + 476.637

shaded area = 676.136… 675.637… = 676

shaded area = 676 [676, 677]

3e. [4 marks]Angle BOC is 2.4 radians.

The shaded region is to be painted red. Red paint is sold in cans which cost each. One can covers . How much does it cost tobuy the paint?

$32 140 m2

Markschemedividing to find number of cans (M1)

e.g. ,

5 cans must be purchased (A1)

multiplying to find cost of cans (M1)

e.g. ,

cost is 160 (dollars) A1 N3

[4 marks]

Examiners reportMost candidates had a suitable strategy for part (e) and knew to work with a whole number of cans of paint.

676140

4.82857…

5(32) × 32676140

4a. [3 marks]

The following diagram shows the graph of , for .

There is a minimum point at P(2, − 3) and a maximum point at Q(4, 3) .

(i) Write down the value of a .

(ii) Find the value of b .

f(x) = acos(bx) 0 ≤ x ≤ 4

Markscheme(i) A1 N1

(ii) METHOD 1

attempt to find period (M1)

e.g. 4 , ,

A1 N2

[3 marks]

METHOD 2

attempt to substitute coordinates (M1)

e.g. ,

A1 N2

[3 marks]

Examiners reportIn part (a), many candidates were able to successfully write down the value of a as instructed by inspecting the graph and seeing theamplitude of the function is 3. Many also used a formulaic approach to reach the correct answer. When finding the value of b, therewere many candidates who thought b was the period of the function, rather than .

a = 3

b = 4 2π

b

b = (= )2π

4π

2

3 cos(2b) = −3 3 cos(4b) = 3

b = (= )2π

4π

2

2π

period

4b. [1 mark]Write down the gradient of the curve at P.

Markscheme0 A1 N1

[1 mark]

Examiners reportIn part (b), the directions asked candidates to write down the gradient of the curve at the local minimum point P. However, manycandidates spent a good deal of time finding the derivative of the function and finding the value of the derivative for the given valueof x, rather than simply stating that the gradient of a curve at a minimum point is zero.

4c. [2 marks]Write down the equation of the normal to the curve at P.

Markschemerecognizing that normal is perpendicular to tangent (M1)

e.g. , , sketch of vertical line on diagram

(do not accept 2 or ) A1 N2

[2 marks]

Examiners reportFor part (c), finding the equation of the normal to the curve, many candidates tried to work with algebraic equations involvingnegative reciprocal gradients, rather than recognizing that the equation of the vertical line was . There were also candidateswho had trouble expressing the correct equation of a line parallel to the y-axis.

× = −1m1 m2 m = − 10

x = 2 y = 2

x = 2

f(x) = acos(b(x − c)) − 1

5a. [2 marks]

The diagram below shows part of the graph of , where .

The point is a maximum point and the point is a minimum point.

Find the value of a .

Markschemeevidence of valid approach (M1)

e.g. , distance from

A1 N2

[2 marks]

Examiners reportA pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph.

f(x) = acos(b(x − c)) − 1 a > 0

P ( ,2)π

4Q( ,−4)3π

4

max y value−min y value2

y = −1

a = 3

5b. [4 marks](i) Show that the period of f is .

(ii) Hence, find the value of b .

Markscheme(i) evidence of valid approach (M1)

e.g. finding difference in x-coordinates,

evidence of doubling A1

e.g.

AG N0

(ii) evidence of valid approach (M1)

e.g.

A1 N2

[4 marks]

π

π

2

2 × ( )π

2

period = π

b = 2π

π

b = 2

Examiners reportA pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph. Some candidates had troubleshowing that the period was , either incorrectly adding the given and or using the value of b that they found first for part(b)(ii).

π π/4 3π/4

5c. [1 mark]Given that , write down the value of c .

Markscheme A1 N1

[1 mark]

Examiners reportA pleasing number of candidates correctly found the values of a, b, and c for this sinusoidal graph. Some candidates had troubleshowing that the period was , either incorrectly adding the given and or using the value of b that they found first for part(b)(ii).

0 < c < π

c = π

4

π π/4 π/3

6a. [2 marks]

Let .

Show that can be expressed as .

Markschemeattempt to expand (M1)

e.g. ; at least 3 terms

correct expansion A1

e.g.

AG N0

[2 marks]

Examiners reportSimplifying a trigonometric expression and applying identities was generally well answered in part (a), although some candidateswere certainly helped by the fact that it was a "show that" question.

f(x) = (sin x + cosx)2

f(x) 1 + sin 2x

(sin x + cosx)(sin x + cosx)

x + 2 sin xcosx + xsin2 cos2

f(x) = 1 + sin 2x

6b. [2 marks]The graph of f is shown below for .

Let . On the same set of axes, sketch the graph of g for .

Markscheme

A1A1 N2

Note: Award A1 for correct sinusoidal shape with period and range , A1 for minimum in circle.

Examiners reportMore candidates had difficulty with part (b) with many assuming the first graph was and hence sketching a horizontaltranslation of for the graph of g; some attempts were not even sinusoidal. While some candidates found the stretch factor pcorrectly or from follow-through on their own graph, very few successfully found the value and direction for the translation.

0 ≤ x ≤ 2π

g(x) = 1 + cosx 0 ≤ x ≤ 2π

2π [0, 2]

1 + sin(x)π/2

6c. [2 marks]The graph of g can be obtained from the graph of f under a horizontal stretch of scale factor p followed by a translation by the

vector .

Write down the value of p and a possible value of k .

Markscheme , A1A1 N2

[2 marks]

( )k

0

p = 2 k = − π

2

Examiners reportPart (c) certainly served as a discriminator between the grade 6 and 7 candidates.

7a. [3 marks]

The diagram shows a circle of radius metres. The points ABCD lie on the circumference of the circle.

BC = m, CD = m, AD = m, , and .

Find AC.

Markschemeevidence of choosing cosine rule (M1)

eg ,


eg ,

AC (m) A1 N2

[3 marks]

Examiners reportThere was an error on this question, where the measurements were inconsistent. Whichever method a candidate used to answer thequestion, the inconsistencies did not cause a problem. The markscheme included a variety of solutions based on possiblecombinations of solutions, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected.Candidate scripts did not indicate any adverse effect.

Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended.Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due tothe fact that cyclic quadrilaterals is not part of the syllabus.

8

14 11.5 8 A C =D 104∘ B D =C 73∘

= + − 2ab cosCc2 a2 b2 C + A − 2 × CD × ADcosDD2 D2

+ − 2 × 11.5 × 8 cos10411.52 82 196.25 − 184 cos104

= 15.5

7b. [5 marks](i) Find .

(ii) Hence, find .

A DC

A BC

Markscheme(i) METHOD 1

evidence of choosing sine rule (M1)

eg ,


eg

A1 N2

METHOD 2

evidence of choosing cosine rule (M1)

eg


e.g.

A1 N2

(ii) subtracting their from (M1)

eg ,

A1 N2

[5 marks]


Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended.Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due tothe fact that cyclic quadrilaterals is not part of the syllabus.

=sin A

a

sin B

b=sin A DC

ADsin D

AC

=sin A DC8

sin 10415.516…

A D =C 30.0∘

= + − 2ab cosCc2 a2 b2

= + 15.516 − 2(11.5)(15.516…)cosC82 11.52 …2

A D =C 30.0∘

A DC 73

73 − A DC 70 − 30.017…

A B =C 43.0∘

7c. [6 marks](c) Find the area of triangle ADC.

(d) Hence or otherwise, find the total area of the shaded regions.

Markscheme(c) correct substitution (A1)

eg area

area (m ) A1 N2

[2 marks]

(d) attempt to subtract (M1)

eg ,

area (A1)

correct working A1

eg ,

shaded area is (m ) A1 N3

[4 marks]

Total [6 marks]


Despite this unfortunate error, the question posed few difficulties for candidates and most approached the problem as intended.Although there were other ways to approach the problem (using properties of cyclic quadrilaterals) few considered this, likely due tothe fact that cyclic quadrilaterals is not part of the syllabus. Candidates were proficient in their use of sine and cosine rules and mostcould find the area of the required triangle in part (c). Those who made errors in this question either had their GDC in the wrongmode or were rounding values prematurely while some misinformed candidates treated ADC as a right-angled triangle. In part (d),most candidates recognized what to do and often obtained follow through marks from errors made in previous parts.

ΔADC = (8)(11.5)sin 10412

= 44.6 2

circle − ABCD π − ΔADC − ΔACBr2

ΔACB = (15.516…)(14)sin 42.9812

π(8 − 44.6336… − (15.516…)(14)sin 42.98)2 12

64π − 44.6 − 74.1

82.4 2

8a. [2 marks]

A Ferris wheel with diameter metres rotates clockwise at a constant speed. The wheel completes rotations every hour. The bottom ofthe wheel is metres above the ground.

A seat starts at the bottom of the wheel.

Find the maximum height above the ground of the seat.

122 2.413


eg ,

maximum height (m) A1 N2

[2 marks]

Examiners reportMost candidates were successful with part (a).

13 + diameter 13 + 122

= 135

8b. [9 marks]

After t minutes, the height metres above the ground of the seat is given by

(b) (i) Show that the period of is minutes.

(ii) Write down the exact value of .

(c) Find the value of .

(d) Sketch the graph of , for .

h

h = 74 + acosbt.

h 25

b

a

h 0 ≤ t ≤ 50

Markscheme(a) (i) period A1

period minutes AG N0

(ii) A1 N1

[2 marks]

(b) METHOD 1

valid approach (M1)

eg , ,

(accept ) (A1)

A1 N2

METHOD 2

attempt to substitute valid point into equation for h (M1)

eg

correct equation (A1)

eg ,

A1 N2

[3 marks]

(c)

A1A1A1A1 N4

Note: Award A1 for approximately correct domain, A1 for approximately correct range,

A1 for approximately correct sinusoidal shape with cycles.

Only if this last A1 awarded, award A1 for max/min in approximately correct positions.

[4 marks]

Total [9 marks]

Examiners reportA surprising number had difficulty producing enough work to show that the period was ; writing down the exactvalue of also overwhelmed a number of candidates. In part (c), candidates did not recognize that the seat on theFerris wheel is a minimum at thereby making the value of a negative. Incorrect values of were often seenwith correct follow through obtained when sketching the graph in part (d). Graphs again frequently failed to show keyfeatures in approximately correct locations and candidates lost marks for incorrect domains and ranges.

= 602.4

= 25

b = 2π

25(= 0.08π)

max − 74 |a| = 135−132

74 − 13

|a| = 61 a = 61

a = −61

135 = 74 + acos( )2π×12.525

135 = 74 + acos(π) 13 = 74 + a

a = −61

2

25b

t = 0 61

8c. [5 marks]In one rotation of the wheel, find the probability that a randomly selected seat is at least metres above the ground.105

Markschemesetting up inequality (accept equation) (M1)

eg , , sketch of graph with line

any two correct values for t (seen anywhere) A1A1

eg , , ,

valid approach M1

eg , , ,

A1 N2

[5 marks]

Examiners reportPart (e) was very poorly done for those who attempted the question and most did not make the connection between height, time andprobability. The idea of linking probability with a real-life scenario proved beyond most candidates. That said, there were a few novelapproaches from the strongest of candidates using circles and angles to solve this part of question 10.

h > 105 105 = 74 + acosbt y = 105

t = 8.371… t = 16.628… t = 33.371… t = 41.628…

16.628−8.37125

−t1 t2

252×8.257

502(12.5−8.371)

25

p = 0.330

9. [6 marks]

The diagram below shows part of the graph of a function .

The graph has a maximum at A( , ) and a minimum at B( , ) .

The function can be written in the form . Find the value of

(a)

(b)

(c) .

f

1 5 3 −1

f f(x) = p sin(qx) + r

p

q

r

Markscheme(a) valid approach to find (M1)

eg amplitude ,

A1 N2

[2 marks]

(b) valid approach to find (M1)

eg period = 4 ,

A1 N2

[2 marks]

(c) valid approach to find (M1)

eg axis = , sketch of horizontal axis,

A1 N2

[2 marks]

Total [6 marks]

Examiners reportMany candidates were able to answer all three parts of this question with no difficulty. Some candidates ran into problems when theyattempted to substitute into the equation of the function with the parameters , and . The successful candidates were able to find theanswers using the given points and their understanding of the different transformations.Part (b) seemed to be the most difficult, with some candidates not understanding the relationship between and the period of thefunction. There were also some candidates who showed working such as without explaining what represented.

p

= max−min2

p = 6

p = 3

q

q = 2π

period

q = π

2

r

max+min2

f(0)

r = 2

p q r

q2π

bb

10a. [3 marks]

The following diagram shows a triangle ABC.

The area of triangle ABC is cm , AB cm , AC cm and .

Find .

80 2 = 18 = x B C =A 50∘

x

Markschemecorrect substitution into area formula (A1)

eg

setting their area expression equal to (M1)

eg

A1 N2

[3 marks]

Examiners reportThe vast majority of candidates were very successful with this question. A small minority drew an altitude from C and used righttriangle trigonometry. Errors included working in radian mode, assuming that the angle at C was , and incorrectly applying theorder of operations when evaluating the cosine rule.

(18x)sin 5012

80

9x sin 50 = 80

x = 11.6

90∘

10b. [3 marks]Find BC.


eg

correct substitution into right hand side (may be in terms of ) (A1)

eg

BC A1 N2

[3 marks]

Examiners reportThe vast majority of candidates were very successful with this question. A small minority drew an altitude from C and used righttriangle trigonometry. Errors included working in radian mode, assuming that the angle at C was , and incorrectly applying theorder of operations when evaluating the cosine rule.

= + + 2ab sin Cc2 a2 b2

x

+ − 2(11.6)(18)cos5011.62 182

= 13.8

90∘

11. [7 marks]

The following diagram shows a circle with centre O and radius cm.

Points A and B are on the circumference of the circle and radians .

The point C is on [OA] such that radians .

The area of the shaded region is cm . Find the value of .

r

A B = 1.4O

B O =C π

2

25 2 r

Markschemecorrect value for BC

eg , (A1)

area of A1

area of sector A1

attempt to subtract in any order (M1)

eg sector – triangle,

correct equation A1

eg

attempt to solve their equation (M1)

eg sketch, writing as quadratic,

A1 N4

[7 marks]

Note: Exception to FT rule. Award A1FT for a correct FT answer from a quadratic equation involving two trigonometric functions.

Examiners reportAs to be expected, candidates found this problem challenging. Those who used a systematic approach in part (b) were moresuccessful than those whose work was scattered about the page. While a pleasing number of candidates successfully found the area ofsector AOB, far fewer were able to find the area of triangle BOC. Candidates who took an analytic approach to solving the resultingequation were generally less successful than those who used their GDC. Candidates who converted the angle to degrees generallywere not very successful.

BC = rsin 1.4 −r2 (rcos1.4)2− −−−−−−−−−−−√

ΔOBC = rsin 1.4 × rcos1.412

(= sin 1.4 × cos1.4)12

r2

OAB = × 1.412

r2

sin 1.4 × cos1.4 − 0.712

r2 r2

0.7 − rsin 1.4 × rcos1.4 = 25r2 12

250.616…

r = 6.37

12a. [1 mark]

The following diagram shows , where RQ = 9 cm, and .

Find .

Markscheme A1 N1

[1 mark]

Examiners reportThis question was attempted in a satisfactory manner.

ΔPQR P Q =R 70∘ P R =Q 45∘

R QP

R Q =P 65∘

12b. [3 marks]Find PR .

Markschemeevidence of choosing sine rule (M1)


e.g.

7.021854078

A1 N2

[3 marks]

Examiners reportThe sine rule was applied satisfactory in part (b) but some obtained an incorrect answer due to having their calculators in radianmode. Some incorrect substitutions were seen, either by choosing an incorrect side or substituting 70 instead of . Approachesusing a combination of the cosine rule and/or right-angled triangle trigonometry were seen.

=PRsin 45∘

9sin 65∘

PR = 7.02

sin 70∘

12c. [2 marks]Find the area of .

Markschemecorrect substitution (A1)

e.g.

A1 N2

[2 marks]

Examiners reportApproaches using a combination of the cosine rule and/or right-angled triangle trigonometry were seen, especially in part (c) tocalculate the area of the triangle.

A few candidates set about finding the height, then used the formula for the area of a right-angled triangle.

ΔPQR

area = × 9 × 7.02… × sin12

70∘

29.69273008

area = 29.7

13a. [4 marks]

Consider the following circle with centre O and radius r .

The points P, R and Q are on the circumference, , for .

Use the cosine rule to show that .

P Q = 2θO 0 < θ < π

2

PQ = 2rsin θ

Markschemecorrect substitution into cosine rule A1

e.g. ,

substituting for (seen anywhere) A1

e.g.

working towards answer (A1)

e.g.

recognizing (including crossing out) (seen anywhere)

e.g. ,

AG N0

[4 marks]

Examiners reportThis exercise seemed to be challenging for the great majority of the candidates, in particular parts (b), (c) and (d).

Part (a) was generally attempted using the cosine rule, but many failed to substitute correctly into the right hand side or skippedimportant steps. A high percentage could not arrive at the given expression due to a lack of knowledge of trigonometric identities ormaking algebraic errors, and tried to force their way to the given answer.

The most common errors included taking the square root too soon, and sign errors when distributing the negative after substituting by .

P = + − 2(r)(r)cos(2θ)Q2 r2 r2 P = 2 − 2 (cos(2θ))Q2 r2 r2

1 − 2 θsin2 cos2θ

P = 2 − 2 (1 − 2 θ)Q2 r2 r2 sin2

P = 2 − 2 + 4 θQ2 r2 r2 r2sin2

2 − 2 = 0r2 r2

P = 4 θQ2 r2sin2 PQ = 4 θr2sin2− −−−−−√

PQ = 2rsinθ

cos2θ 1 − 2 θsin2

13b. [5 marks]Let l be the length of the arc PRQ .

Given that , find the value of .

Markscheme (seen anywhere) (A1)

correct set up A1

e.g.

attempt to eliminate r (M1)

correct equation in terms of the one variable (A1)

e.g.

1.221496215

(accept (69.9)) A1 N3

[5 marks]

Examiners reportThis exercise seemed to be challenging for the great majority of the candidates, in particular parts (b), (c) and (d).

In part (b), most candidates understood what was required but could not find the correct length of the arc PRQ mainly due tosubstituting the angle by instead of .

1.3PQ − l = 0 θ

PRQ = r × 2θ

1.3 × 2rsin θ − r × (2θ) = 0

θ

1.3 × 2 sin θ − 2θ = 0

θ = 1.22 70.0∘

θ 2θ

13c. [4 marks]Consider the function , for .

(i) Sketch the graph of f .

(ii) Write down the root of .

f(θ) = 2.6sin θ − 2θ 0 < θ < π

2

f(θ) = 0

Markscheme(i)

A1A1A1 N3

Note: Award A1 for approximately correct shape, A1 for x-intercept in approximately correct position, A1 for domain. Do notpenalise if sketch starts at origin.

(ii)

A1 N1

[4 marks]

Examiners reportRegarding part (c), many valid approaches were seen for the graph of f, making a good use of their GDC. A common error wasfinding a second or third solution outside the domain. A considerable amount of sketches were missing a scale.

There were candidates who achieved the correct equation but failed to realize they could use their GDC to solve it.

1.221496215

θ = 1.22

13d. [3 marks]Use the graph of f to find the values of for which .

Markschemeevidence of appropriate approach (may be seen earlier) M2

e.g. , , showing positive part of sketch

(accept ) A1 N1

[3 marks]

Examiners reportPart (d) was attempted by very few, and of those who achieved the correct answer not many were able to show the method they used.

θ l < 1.3PQ

2θ < 2.6sin θ 0 < f(θ)

0 < θ < 1.221496215

0 < θ = 1.22 θ < 1.22

14a. [4 marks]

The following diagram shows a triangle ABC.

, radians , , , where .

(i) Show that .

(ii) Find p .


e.g. choosing cosine rule

correct substitution (A1)

e.g.

simplification A1

e.g.

AG N0

(ii)

A1 N1

Note: Award A0 for , i.e. not rejecting the negative value.

[4 marks]

Examiners reportThere were mixed results with this question. Most candidates could access part (a) and made the correct choice with the cosine rulebut sloppy notation often led to candidates not being able to show the desired result.

BC = 6 C B = 0.7A AB = 4p AC = 5p p > 0

(41 − 40 cos0.7) = 36p2

= (5p + (4p − 2 × (4p) × (5p)cos0.762 )2 )2

36 = 25 + 16 − 40 cos0.7p2 p2 p2

(41 − 40 cos0.7) = 36p2

1.85995…

p = 1.86

p = ±1.86

A Bˆ

14b. [1 mark]

Consider the circle with centre B that passes through the point C. The circle cuts the line CA at D, and is obtuse. Part of the circle isshown in the following diagram.

Write down the length of BD.

Markscheme A1 N1

[1 mark]

Examiners reportThere were mixed results with this question. Most candidates could access part (a) and made the correct choice with the cosine rulebut sloppy notation often led to candidates not being able to show the desired result.

A BD

BD = 6

14c. [4 marks]Find .

Markschemeevidence of valid approach (M1)

e.g. choosing sine rule


e.g.

(A1)

A1 N3

[4 marks]

Examiners reportIn part (c), candidates again correctly identified an appropriate method but failed to recognize that their result of 0.925 was acute andnot obtuse as required.

A BD

=sin A BD4p

sin 0.76

acute A B = 0.9253166…D

π − 0.9253166… = 2.216275…

A B = 2.22D

14d. [6 marks](i) Show that radians, correct to 2 decimal places.

(ii) Hence, find the area of the shaded region.

C D = 1.29B


e.g. recognize isosceles triangle, base angles equal

A1

AG N0

(ii) area of sector BCD (A1)

e.g.

area of triangle BCD (A1)

e.g.

evidence of subtraction M1

A1 N3

[6 marks]

Examiners reportIn (d) (i), many attempted to use the sine rule under the incorrect assumption that DC was equal to 5p, rather than rely on some basicisosceles triangle geometry. Consequently, the result of 1.29 for was not easy to show. There was a great deal of success with(d) (ii) with candidates using appropriate techniques to find the area of the shaded region although some stopped after finding the areaof the sector.

π − 2(0.9253…)

C D = 1.29B

0.5 × (1.29) × (6)2

0.5 × (6 sin 1.29)2

5.92496…

5.937459…

area = 5.94

C DB

15a. [3 marks]

Let , where .

Find .

MarkschemeMETHOD 1

evidence of choosing (M1)


e.g. , ,

A1 N2

Note: If no working shown, award N1 for .

METHOD 2

approach involving Pythagoras’ theorem (M1)

e.g. ,

finding third side equals 3 (A1)

A1 N2

Note: If no working shown, award N1 for .

[3 marks]

sin θ = 213√

< θ < ππ

2

cosθ

θ + θ = 1sin2 cos2

θ =cos2 913

cosθ = ± 313√

cosθ = 913

−−√cosθ = − 3

13√

313√

+ = 1322 x2

cosθ = − 313√

313√

Examiners reportWhile the majority of candidates knew to use the Pythagorean identity in part (a), very few remembered that the cosine of an angle inthe second quadrant will have a negative value.

15b. [5 marks]Find .

Markschemecorrect substitution into (seen anywhere) (A1)

e.g.

correct substitution into (seen anywhere) (A1)

e.g. , ,

valid attempt to find (M1)

e.g. ,

correct working A1

e.g. , ,

A1 N4

Note: If students find answers for which are not in the range , award full FT in (b) for correct FT working shown.

[5 marks]

Examiners reportIn part (b), many candidates incorrectly tried to calculate as , rather than using the double-angle identities.

tan2θ

sin 2θ

2 ( ) (− )213√

313√

cos2θ

−(− )313√

2 ( )213√

22 − 1(− )3

13√

21 − 2( )2

13√

2

tan2θ

2( )(− )2

13√

3

13√

−(− )3

13√

2 ( )2

13√

2

2(− )23

1−(− )23

2

(2)(2)(−3)

13

−913

413

− 12

( )13√ 2

−11813

− 1213

513

tan2θ = − 125

cosθ [−1, 1]

tan2θ 2 × tanθ

f(x) = asin(b(x − c)) + d

16a. [3 marks]

The following diagram shows the graph of , for .

There is a maximum point at P(4, 12) and a minimum point at Q(8, −4) .

Use the graph to write down the value of

(i) a ;

(ii) c ;

(iii) d .

Markscheme(i) A1 N1

(ii) A1 N1

(iii) A1 N1

[3 marks]

Examiners reportPart (a) of this question proved challenging for most candidates.

f(x) = asin(b(x − c)) + d 2 ≤ x ≤ 10

a = 8

c = 2

d = 4

16b. [2 marks]Show that .b = π

4

MarkschemeMETHOD 1

recognizing that period (A1)

correct working A1

e.g. ,

AG N0

METHOD 2

attempt to substitute M1

e.g.

correct working A1

e.g.

AG N0

[2 marks]

Examiners reportAlthough a good number of candidates recognized that the period was 8 in part (b), there were some who did not seem to realize thatthis period could be found using the given coordinates of the maximum and minimum points.

= 8

8 = 2π

bb = 2π

8

b = π

4

12 = 8 sin(b(4 − 2)) + 4

sin 2b = 1

b = π

4

16c. [3 marks]Find .

Markschemeevidence of attempt to differentiate or choosing chain rule (M1)

e.g. ,

(accept ) A2 N3

[3 marks]

Examiners reportIn part (c), not many candidates found the correct derivative using the chain rule.

(x)f ′

cos (x − 2)π

4× 8π

4

(x) = 2πcos( (x − 2))f ′ π

42πcos (x − 2)π

4

16d. [6 marks]At a point R, the gradient is . Find the x-coordinate of R.−2π

Markschemerecognizing that gradient is (M1)

e.g.

correct equation A1

e.g. ,


e.g.

using (seen anywhere) (A1)

e.g.

simplifying (A1)

e.g.

A1 N4

[6 marks]

Examiners reportFor part (d), a good number of candidates correctly set their expression equal to , but errors in their previous values kept mostfrom correctly solving the equation. Most candidates who had the correct equation were able to gain full marks here.

(x)f ′

(x) = mf ′

−2π = 2πcos( (x − 2))π

4−1 = cos( (x − 2))π

4

(−1) = (x − 2)cos−1 π

4

(−1) = πcos−1

π = (x − 2)π

4

4 = (x − 2)

x = 6

−2π

17a. [2 marks]

Consider the following circle with centre O and radius 6.8 cm.

The length of the arc PQR is 8.5 cm.

Find the value of .

Markschemecorrect substitution (A1)

e.g. ,

(accept ) A1 N2

[2 marks]

Examiners reportPart (a) was almost universally done correctly.

θ

8.5 = θ(6.8) θ = 8.56.8

θ = 1.25 71.6∘

17b. [4 marks]Find the area of the shaded region.

MarkschemeMETHOD 1

correct substitution into area formula (seen anywhere) (A1)

e.g. ,

correct substitution into area formula (seen anywhere) (A1)

e.g. , 28.9

valid approach M1

e.g. ; ;

( ) A1 N2

METHOD 2

attempt to find reflex angle (M1)

e.g. ,

correct reflex angle (A1)

( )

correct substitution into area formula A1

e.g.

( ) A1 N2

[4 marks]

Examiners reportMany also had little trouble in part (b), with most subtracting from the circle's area, and a minority using the reflex angle. A fewcandidates worked in degrees, although some of these did so incorrectly by using the radian area formula. Some candidates onlyfound the area of the unshaded sector.

A = π(6.8)2 145.267…

A = (1.25)( )12

6.82

π(6.8 − (1.25)( ))2 12

6.82 145.267… − 28.9 π − sin θr2 12

r2

A = 116 cm2

2π − θ 360 − 1.25

A B = 2π − 1.25O = 5.03318…

A = (5.03318…)( )12

6.82

A = 116 cm2

18a. [4 marks]

Consider the triangle ABC, where AB =10 , BC = 7 and = .

Find the two possible values of .

MarkschemeNote: accept answers given in degrees, and minutes.


e.g.


e.g. ,

, A1A1 N1N1

Note: If candidates only find the acute angle in part (a), award no marks for (b).

[4 marks]

Examiners reportMost candidates were comfortable applying the sine rule, although many were then unable to find the obtuse angle, demonstrating alack of understanding of the ambiguous case. This precluded them from earning marks in part (b). Those who found the obtuse anglegenerally had no difficulty with part (b).

C BA 30∘

A BC

=sin A

a

sin B

b

=sin θ

10sin 30∘

7sin θ = 5

7

A B =C 45.6∘ A B =C 134∘

18b. [2 marks]Hence, find , given that it is acute.

Markschemeattempt to substitute their larger value into angle sum of triangle (M1)

e.g.

A1 N2

[2 marks]

Examiners reportMost candidates were comfortable applying the sine rule, although many were then unable to find the obtuse angle, demonstrating alack of understanding of the ambiguous case. This precluded them from earning marks in part (b). Those who found the obtuse anglegenerally had no difficulty with part (b).

A CB

− (134.415 + )180∘ …∘ 30∘

A C =B 15.6∘

19. [7 marks]Solve the equation , for .

MarkschemeMETHOD 1

using double-angle identity (seen anywhere) A1

e.g. ,

evidence of valid attempt to solve equation (M1)

e.g. ,

, A1A1

, , A1A1A1 N4

METHOD 2

A1A1M1A1

Notes: Award A1 for sketch of , A1 for a sketch of , M1 for at least one intersection point seen, and A1 for 3approximately correct intersection points. Accept sketches drawn outside , even those with more than 3 intersections.

, , A1A1A1 N4

[7 marks]

Examiners reportBy far the most common error was to “cancel” the and find only two of the three solutions. It was disappointing how fewcandidates solved this by setting factors equal to zero. Some candidates wrote all three answers from , which only earnedtwo of the three final marks. On a brighter note, many candidates found the , which showed an appreciation for the period of thefunction as well as the domain restriction. A handful of candidates cleverly sketched both graphs and used the intersections to find thethree solutions.

2 cosx = sin 2x 0 ≤ x ≤ 3π

sin 2x = 2 sin xcosx 2 cosx = 2 sin xcosx

0 = 2 sin xcosx − 2 cosx 2 cosx(1 − sin x) = 0

cosx = 0 sin x = 1

x = π

2x = 3π

2x = 5π

2

sin 2x 2 cosx

[0,3π]

x = π

2x = 3π

2x = 5π

2

cosx

sin x = 15π

2

20a. [3 marks]

The following diagram shows triangle ABC .

AB = 7 cm, BC = 9 cm and .

Find AC .


e.g.


e.g.

A1 N2

[3 marks]

Examiners reportThe majority of candidates were successful with this question. Most correctly used the cosine rule in part (a) and the sine rule in part(b). Some candidates did not check that their GDC was set in degree mode while others treated the triangle as if it were right angled.A large number of candidates were penalized for not leaving their answers exactly or to three significant figures.

A C =B 120∘

+ − 2ab cosCa2 b2

+ − 2(7)(9)cos72 92 120∘

AC = 13.9 (= )193−−−√

20b. [3 marks]Find .B CA

=sin A

BCsin B

AC

=sin A

9sin 12013.9

=A 34.1∘

cos =A A +A −BB2 C2 C2

2(AB)(AC)

cos =A + −72 13.92 92

2(7)(13.9)

=A 34.1∘

Examiners reportThe majority of candidates were successful with this question. Most correctly used the cosine rule in part (a) and the sine rule in part(b). Some candidates did not check that their GDC was set in degree mode while others treated the triangle as if it were right angled.A large number of candidates were penalized for not leaving their answers exactly or to three significant figures.

21a. [2 marks]

The following diagram shows a waterwheel with a bucket. The wheel rotates at a constant rate in an anticlockwise (counter-clockwise)direction.

The diameter of the wheel is 8 metres. The centre of the wheel, A, is 2 metres above the water level. After t seconds, the height of the bucketabove the water level is given by .

Show that .

MarkschemeMETHOD 1

evidence of recognizing the amplitude is the radius (M1)

e.g. amplitude is half the diameter

A1

AG N0

METHOD 2

evidence of recognizing the maximum height (M1)

e.g. ,

correct reasoning

e.g. and has amplitude of 1 A1

AG N0

[2 marks]

Examiners reportParts (a) and (b) were generally well done.

h = asin bt + 2

a = 4

a = 82

a = 4

h = 6 asin bt + 2 = 6

asin bt = 4 sin bt

a = 4

21b. [2 marks]The wheel turns at a rate of one rotation every 30 seconds.

Show that .b = π

15

MarkschemeMETHOD 1

period = 30 (A1)

A1

AG N0

METHOD 2

correct equation (A1)

e.g. ,

A1

AG N0

[2 marks]

Examiners reportParts (a) and (b) were generally well done, however there were several instances of candidates working backwards from the givenanswer in part (b).

b = 2π

30

b = π

15

2 = 4 sin 30b + 2 sin 30b = 0

30b = 2π

b = π

15

21c. [6 marks]In the first rotation, there are two values of t when the bucket is descending at a rate of .

Find these values of t .

Markschemerecognizing (seen anywhere) R1

attempting to solve (M1)

e.g. sketch of , finding

correct work involving A2

e.g. sketch of showing intersection,

, A1A1 N3

[6 marks]

Examiners reportParts (c) and (d) proved to be quite challenging for a large proportion of candidates. Many did not attempt these parts. The mostcommon error was a misinterpretation of the word "descending" where numerous candidates took to be 0.5 instead of but

incorrect derivatives for h were also widespread. The process required to solve for t from the equation

overwhelmed those who attempted algebraic methods. Few could obtain both correct solutions, more had one correct while othersincluded unreasonable values including .

0.5 ms−1

(t) = −0.5h′

h′ h′

h′

h′ −0.5 = cos( t)4π

15π

15

t = 10.6 t = 19.4

(t)h′ −0.5

−0.5 = cos( t)4π

15π

15

t < 0

21d. [4 marks]In the first rotation, there are two values of t when the bucket is descending at a rate of .

Determine whether the bucket is underwater at the second value of t .

0.5 ms−1

MarkschemeMETHOD 1

valid reasoning for their conclusion (seen anywhere) R1

e.g. so underwater; so not underwater

evidence of substituting into h (M1)

e.g. ,

correct calculation A1

e.g.

correct statement A1 N0

e.g. the bucket is underwater, yes

METHOD 2

valid reasoning for their conclusion (seen anywhere) R1

e.g. so underwater; so not underwater

evidence of valid approach (M1)

e.g. solving , graph showing region below x-axis

correct roots A1

e.g. ,

correct statement A1 N0

e.g. the bucket is underwater, yes

[4 marks]

Examiners reportIn part (d), not many understood that the condition for underwater was and had trouble interpreting the meaning of "secondvalue". Many candidates, however, did recover to gain some marks in follow through.

h(t) < 0 h(t) > 0

h(19.4) 4 sin + 219.4π

15

h(19.4) = −1.19

h(t) < 0 h(t) > 0

h(t) = 0

17.5 27.5

h(t) < 0

22. [6 marks]Let . Find .h(x) = 6x

cos x(0)h′

MarkschemeMETHOD 1 (quotient)

derivative of numerator is 6 (A1)

derivative of denominator is (A1)

attempt to substitute into quotient rule (M1)


e.g.

substituting (A1)

e.g.

A1 N2

METHOD 2 (product)

derivative of 6x is 6 (A1)

derivative of is (A1)

attempt to substitute into product rule (M1)


e.g.

substituting (A1)

e.g.

A1 N2

[6 marks]

Examiners reportThe majority of candidates were successful in using the quotient rule, and were able to earn most of the marks for this question.However, there were a large number of candidates who substituted correctly into the quotient rule, but then went on to make mistakesin simplifying this expression. These algebraic errors kept the candidates from earning the final mark for the correct answer. A fewcandidates tried to use the product rule to find the derivative, but they were generally not as successful as those who used the quotientrule. It was pleasing to note that most candidates did know the correct values for the sine and cosine of zero.

−sin x

(cos x)(6)−(6x)(− sin x)

(cos x)2

x = 0

(cos 0)(6)−(6×0)(− sin 0)

(cos 0)2

(0) = 6h′

h(x) = 6x × (cosx)−1

(cosx)−1 (−(cosx (−sin x)))−2

(6x)(−(cosx (−sin x)) + (6)(cosx)−2 )−1

x = 0

(6 × 0)(−(cos0 (−sin 0)) + (6)(cos0)−2 )−1

(0) = 6h′

23a. [2 marks]

The following diagram represents a large Ferris wheel, with a diameter of 100 metres.

Let P be a point on the wheel. The wheel starts with P at the lowest point, at ground level. The wheel rotates at a constant rate, in ananticlockwise (counter-clockwise) direction. One revolution takes 20 minutes.

Write down the height of P above ground level after

(i) 10 minutes;

(ii) 15 minutes.

Markscheme(i) 100 (metres) A1 N1

(ii) 50 (metres) A1 N1

[2 marks]

Examiners reportNearly all candidates answered part (a) correctly, finding the height of the wheel at and of a revolution.1

234

23b. [4 marks]

Let metres be the height of P above ground level after t minutes. Some values of are given in the table below.

(i) Show that .

(ii) Find .

h(t) h(t)

h(8) = 90.5

h(21)

Markscheme(i) identifying symmetry with (M1)

subtraction A1

e.g. ,

AG N0

(ii) recognizing period (M1)

e.g.

A1 N2

[4 marks]

Examiners reportWhile many candidates were successful in part (b), there were many who tried to use right-angled triangles or find a function forheight, rather than recognizing the symmetry of the wheel in its different positions and using the values given in the table.

h(2) = 9.5

100 − h(2) 100 − 9.5

h(8) = 90.5

h(21) = h(1)

h(21) = 2.4

23c. [3 marks]Sketch the graph of h , for .

Markscheme

A1A1A1 N3

Note: Award A1 for end points (0, 0) and (40, 0) , A1 for range , A1 for approximately correct sinusoidal shape, withtwo cycles.

[3 marks]

Examiners reportIn part (c), most candidates were able to sketch a somewhat accurate representation of the height of the wheel over two full cycles.However, it seems that many candidates are not familiar with the shape of a sinusoidal wave, as many of the candidates' graphs wereconstructed of line segments, rather than a curve.

0 ≤ t ≤ 40

0 ≤ h ≤ 100

23d. [5 marks]Given that h can be expressed in the form , find a , b and c .h(t) = acosbt + c

Markschemeevidence of a quotient involving 20, or to find b (M1)

e.g. ,

(accept if working in degrees) A1 N2

, A2A1 N3

[5 marks]

Examiners reportFor part (d), candidates were less successful in finding the parameters of the cosine function. Even candidates who drew accuratesketches were not always able to relate their sketch to the function. These candidates understood the context of the problem, that theposition on the wheel goes up and down, but they did not relate this to a trigonometric function. Only a small number of candidatesrecognized that the value of a would be negative. Candidates should be aware that while working in degrees may be acceptable, theexpectation is that radians will be used in these types of questions.

2π 360∘

= 202π

bb = 360

20

b = 2π

20(= )π

10b = 18

a = −50 c = 50

24a. [2 marks]

The diagram shows two concentric circles with centre O.

The radius of the smaller circle is 8 cm and the radius of the larger circle is 10 cm.

Points A, B and C are on the circumference of the larger circle such that is radians.

Find the length of the arc ACB .

Markschemecorrect substitution in (A1)

e.g. ,

arc length A1 N2

[2 marks]

Examiners reportThis question was very well done by the majority of candidates. Some candidates correctly substituted the values into the formulas,but failed to do the calculations and write their answers in finished form.

A BO π

3

l = rθ

10 × π

3× 2π × 101

6

= 20π

6(= )10π

3


Markschemearea of large sector (A1)

area of small sector (A1)

evidence of valid approach (seen anywhere) M1

e.g. subtracting areas of two sectors,

area shaded (accept , etc.) A1 N3

[4 marks]

Examiners reportNearly all used the correct method of subtracting the sector areas in part (b), though multiplying with fractions proved challenging forsome candidates.

= × ×12

102 π

3(= )100π

6

= × ×12

82 π

3(= )64π

6

× ( − )12

π

3102 82

= 6π 36π

6

25a. [2 marks]Show that .

Markschemeattempt to substitute for (M1)


e.g.

AG N0

[2 marks]

Examiners reportIn part (a), most candidates successfully substituted using the double-angle formula for cosine. There were quite a few candidateswho worked backward, starting with the required answer and manipulating the equation in various ways. As this was a "show that"question, working backward from the given answer is not a valid method.

4 − cos2θ + 5 sin θ = 2 θ + 5 sin θ + 3sin2

1 − 2 θsin2 cos2θ

4 − (1 − 2 θ) + 5 sin θsin2

4 − cos2θ + 5 sin θ = 2 θ + 5 sin θ + 3sin2

25b. [5 marks]Hence, solve the equation for .

Markschemeevidence of appropriate approach to solve (M1)

e.g. factorizing, quadratic formula

correct working A1

e.g. , ,

correct solution (do not penalise for including (A1)

A2 N3

[5 marks]

Examiners reportIn part (b), many candidates seemed to realize what was required by the word “hence”, though some had trouble factoring thequadratic-type equation. A few candidates were also successful using the quadratic formula. Some candidates got the wrong solutionto the equation , and there were a few who did not realize that the equation has no solution.

4 − cos2θ + 5 sin θ = 0 0 ≤ θ ≤ 2π

(2 sin θ + 3)(sin θ + 1) (2x + 3)(x + 1) = 0 sin x = −5± 1√4

sin θ = −1 sin θ = − 32

θ = 3π

2

sin θ = −1 sin θ = − 32

26a. [1 mark]

The straight line with equation makes an acute angle with the x-axis.

Write down the value of .

Markscheme (do not accept ) A1 N1

[1 mark]

Examiners reportMany candidates drew a diagram to correctly find , although few recognized that a line through the origin can be expressed as

, with gradient , which is explicit in the syllabus.

y = x34

θ

tanθ

tanθ = 34

x34

tanθ

y = x tanθ tanθ

26b. [6 marks]Find the value of

(i) ;

(ii) .

Markscheme(i) , (A1)(A1)


e.g.

A1 N3

(ii) correct substitution A1

e.g. ,

A1 N1

[6 marks]

Examiners reportA surprising number were unable to find the ratios for and from . It was not uncommon for candidates to useunreasonable values, such as and , or to write nonsense such as .

sin 2θ

cos2θ

sin θ = 35

cosθ = 45

sin 2θ = 2 ( ) ( )35

45

sin 2θ = 2425

cos2θ = 1 − 2( )35

2−( )4

5

2 ( )35

2

cos2θ = 725

sin θ cosθ tanθ

sin θ = 3 cosθ = 4 2 sin cos35

45

27a. [5 marks]

Let , for .

Use the quotient rule to show that .

f(x) = cos x

sin xsin x ≠ 0

(x) =f ′ −1xsin2

Markscheme , (seen anywhere) (A1)(A1)

evidence of using the quotient rule M1


e.g. ,

A1

AG N0

[5 marks]

Examiners reportMany candidates comfortably applied the quotient rule, although some did not completely show that the Pythagorean identityachieves the numerator of the answer given. Whether changing to , or applying the quotient rule a second time, mostcandidates neglected the chain rule in finding the second derivative.

sin x = cosxddx

cosx = −sin xddx

sin x(− sin x)−cos x(cos x)

xsin2

− x− xsin2 cos2

xsin2

(x) =f ′ −( x+ x)sin2 cos2

xsin2

(x) =f ′ −1xsin2

−(sin x)−2


MarkschemeMETHOD 1

appropriate approach (M1)

e.g.

A1A1 N3

Note: Award A1 for , A1 for .

METHOD 2

derivative of (seen anywhere) A1

evidence of choosing quotient rule (M1)

e.g. , ,

A1 N3

[3 marks]

Examiners reportWhether changing to , or applying the quotient rule a second time, most candidates neglected the chain rule in finding thesecond derivative.

(x)f ′′

(x) = −(sin xf ′ )−2

(x) = 2( x)(cosx)f ′′ sin−3 (= )2 cos x

xsin3

2 xsin−3 cosx

x = 2 sin xcosxsin2

u = −1 v = xsin2 =f ′′ x×0−(−1)2 sin x cos xsin2

( x)sin2 2

(x) =f ′′ 2 sin x cos x

( x)sin2 2(= )2 cos x

xsin3

−(sin x)−2

( ) ( )

27c. [3 marks]

In the following table, and . The table also gives approximate values of and near .

Find the value of p and of q.

Markschemeevidence of substituting M1

e.g. ,

, A1A1 N1N1

[3 marks]

Examiners reportThose who knew the trigonometric ratios at typically found the values of p and of q, sometimes in follow-through from an incorrect

.

( ) = pf ′ π

2( ) = qf ′′ π

2(x)f ′ (x)f ′′ x = π

2

π

2

−1sin2 π

2

2 cos π

2

sin3 π

2

p = −1 q = 0

π

2(x)f ′′

27d. [2 marks]Use information from the table to explain why there is a point of inflexion on the graph of f where .

Markschemesecond derivative is zero, second derivative changes sign R1R1 N2

[2 marks]

Examiners reportFew candidates gave two reasons from the table that supported the existence of a point of inflexion. Most stated that the secondderivative is zero and neglected to consider the sign change to the left and right of q. Some discussed a change of concavity, butwithout supporting this statement by referencing the change of sign in , so no marks were earned.

x = π

2

(x)f ′′

28a. [2 marks]

Let and .

Find .

Markscheme (A1)

A1 N2

[2 marks]

f(x) = cos2x g(x) = 2 − 1x2

f ( )π

2

f ( ) = cosππ

2

= −1

Examiners reportIn part (a), a number of candidates were not able to evaluate , either leaving it or evaluating it incorrectly.cosπ


Markscheme (A1)

A1 N2

[2 marks]

Examiners reportAlmost all candidates evaluated the composite function in part (b) in the given order, many earning follow-through marks forincorrect answers from part (a). On both parts (a) and (b), there were candidates who correctly used double-angle formulas to comeup with correct answers; while this is a valid method, it required unnecessary additional work.

(g ∘ f)( )π

2

(g ∘ f)( ) = g(−1)π

2(= 2(−1 − 1))2

= 1

28c. [3 marks]Given that can be written as , find the value of k, .

Markscheme A1

evidence of (seen anywhere) (M1)

A1 N2

[3 marks]

Examiners reportCandidates were not as successful in part (c). Many tried to use double-angle formulas, but either used the formula incorrectly or usedit to write the expression in terms of and went no further. There were a number of cases in which the candidates "accidentally"came up with the correct answer based on errors or lucky guesses and did not earn credit for their final answer. Only a few candidatesrecognized the correct method of solution.

(g ∘ f)(x) cos(kx) k ∈ Z

(g ∘ f)(x) = 2(cos(2x) − 1)2 (= 2 (2x) − 1)cos2

2 θ − 1 = cos2θcos2

(g ∘ f)(x) = cos4x

k = 4

cosx

f(x) = 6 + 6 sin x

29a. [5 marks]

Let . Part of the graph of f is shown below.

The shaded region is enclosed by the curve of f , the x-axis, and the y-axis.

Solve for

(i) ;

(ii) .

Markscheme(i) A1

, A1A1 N2

(ii) A1

A1 N1

[5 marks]

Examiners reportMany candidates again had difficulty finding the common angles in the trigonometric equations. In part (a), some did not showsufficient working in solving the equations. Others obtained a single solution in (a)(i) and did not find another. Some candidatesworked in degrees; the majority worked in radians.

f(x) = 6 + 6 sin x

0 ≤ x < 2π

6 + 6 sin x = 6

6 + 6 sin x = 0

sin x = 0

x = 0 x = π

sin x = −1

x = 3π

2

29b. [1 mark]Write down the exact value of the x-intercept of f , for .

Markscheme A1 N1

[1 mark]

Examiners reportWhile some candidates appeared to use their understanding of the graph of the original function to find the x-intercept in part (b), mostused their working from part (a)(ii) sometimes with follow-through on an incorrect answer.

0 ≤ x < 2π

3π

2

29c. [6 marks]The area of the shaded region is k . Find the value of k , giving your answer in terms of .π

Markschemeevidence of using anti-differentiation (M1)

e.g.

correct integral (seen anywhere) A1A1


e.g. ,

A1A1 N3

[6 marks]

Examiners reportMost candidates recognized the need for integration in part (c) but far fewer were able to see the solution through correctly to the end.Some did not show the full substitution of the limits, having incorrectly assumed that evaluating the integral at 0 would be 0; withoutthis working, the mark for evaluating at the limits could not be earned. Again, many candidates had trouble working with thecommon trigonometric values.

(6 + 6 sin x)dx∫3π

20

6x − 6 cosx

6 ( ) − 6 cos( ) − (−6 cos0)3π

23π

29π − 0 + 6

k = 9π + 6

29d. [2 marks]Let . The graph of f is transformed to the graph of g.

Give a full geometric description of this transformation.

Markschemetranslation of A1A1 N2

[2 marks]

Examiners reportWhile there was an issue in the wording of the question with the given domains, this did not appear to bother candidates in part (d).This part was often well completed with candidates using a variety of language to describe the horizontal translation to the right by .

g(x) = 6 + 6 sin (x − )π

2

( )π

2

0

π

2

29e. [3 marks]Let . The graph of f is transformed to the graph of g.

Given that and , write down the two values of p.

Markschemerecognizing that the area under g is the same as the shaded region in f (M1)

, A1A1 N3

[3 marks]

Examiners reportMost candidates who attempted part (e) realized that the integral was equal to the value that they had found in part (c), but a majoritytried to integrate the function g without success. Some candidates used sketches to find one or both values for p. The problem in thewording of the question did not appear to have been noticed by candidates in this part either.

g(x) = 6 + 6 sin (x − )π

2

g(x)dx = k∫ p+ 3π

2p

0 ≤ p < 2π

p = π

2p = 0

30a. [4 marks]

The following diagram shows the triangle ABC.

The angle at C is obtuse, , and the area is .

Find .

Markschemecorrect substitution into the formula for the area of a triangle A1

e.g. ,

attempt to solve (M1)

e.g. ,

( ) (A1)

A1 N3

[4 marks]

Examiners reportPart (a) was well done with the majority of candidates obtaining the acute angle. Unfortunately, the question asked for the obtuseangle which was clearly stated and shown in the diagram. No matter which angle was used, most candidates were able to obtain fullmarks in part (b) with a simple application of the cosine rule.

AC = 5 cm BC = 13.6 cm 20 cm2

A BC

× 5 × 13.6 × sin C = 2012

× 5 × h = 2012

sin C = 0.5882… sin C = 813.6

= 36.031C …∘ 0.6288… radians

A B =C 144∘ (2.51 radians)

30b. [3 marks]Find AB.

Markschemeevidence of choosing the cosine rule (M1)


e.g.

A1 N2

[3 marks]

Examiners reportPart (a) was well done with the majority of candidates obtaining the acute angle. Unfortunately, the question asked for the obtuseangle which was clearly stated and shown in the diagram. No matter which angle was used, most candidates were able to obtain fullmarks in part (b) with a simple application of the cosine rule.

(AB = + − 2(5)(13.6)cos143.968…)2 52 13.62

AB = 17.9

31a. [2 marks]

The following diagram represents a large Ferris wheel at an amusement park.

The points P, Q and R represent different positions of a seat on the wheel.

The wheel has a radius of 50 metres and rotates clockwise at a rate of one revolution every 30 minutes.

A seat starts at the lowest point P, when its height is one metre above the ground.

Find the height of a seat above the ground after 15 minutes.


e.g. 15 mins is half way, top of the wheel,

height (metres) A1 N2

[2 marks]

Examiners reportPart (a) was well done with most candidates obtaining the correct answer.

d + 1

= 101

31b. [5 marks]After six minutes, the seat is at point Q. Find its height above the ground at Q.

Markschemeevidence of identifying rotation angle after 6 minutes A1

e.g. , of a rotation,

evidence of appropriate approach (M1)

e.g. drawing a right triangle and using cosine ratio

correct working (seen anywhere) A1

e.g. ,

evidence of appropriate method M1

e.g. height

height (metres) (accept 35.6) A1 N2

[5 marks]

2π

515

72∘

cos =2π

5x

5015.4(508 …)

= radius + 1 − 15.45…

= 35.5

Examiners reportPart (b) however was problematic with most errors resulting from incorrect, missing or poorly drawn diagrams. Many did notrecognize this as a triangle trigonometric problem while others used the law of cosines to find the chord length rather than the verticalheight, but this was only valid if they then used this to complete the problem. Many candidates misinterpreted the question as one thatwas testing arc length and area of a sector and made little to no progress in part (b).

Still, others recognized that 6 minutes represented of a rotation, but the majority then thought the height after 6 minutes should be of the maximum height, treating the situation as linear. There were even a few candidates who used information given later in thequestion to answer part (b). Full marks are not usually awarded for this approach.

15

15

31c. [6 marks]The height of the seat above ground after t minutes can be modelled by the function .

Find the value of b and of c .

MarkschemeMETHOD 1

evidence of substituting into (M1)

correct substitution

e.g. period = 30 minutes, A1

A1 N2

substituting into (M1)

e.g. ,


A1 N2

METHOD 2

evidence of setting up a system of equations (M1)

two correct equations

e.g. , A1A1

attempt to solve simultaneously (M1)

e.g. evidence of combining two equations

, A1A1 N2N2

[6 marks]

Examiners reportPart (c) was not well done. It was expected that candidates simply use the formula to find the value of b and then substitute backinto the equation to find the value of c. However, candidates often preferred to set up a pair of equations and attempt to solve themanalytically, some successful, some not. No attempts were made to solve this system on the GDC indicating that candidates do not getexposed to many “systems” that are not linear. Confusing radians and degrees here did nothing to improve the lack of success.

h(t) = 50 sin(b(t − c)) + 51

b = 2π

period

b = 2π

30

b = 0.209 ( )π

15

h(t)

h(0) = 1 h(15) = 101

1 = 50 sin (− c) + 51π

15

c = 7.5

1 = 50 sin b(0 − c) + 51 101 = 50 sin b(15 − c) + 51

b = 0.209 ( )π

15c = 7.5

2π

period

31d. [3 marks]The height of the seat above ground after t minutes can be modelled by the function .

Hence find the value of t the first time the seat is above the ground.

h(t) = 50 sin(b(t − c)) + 51

96 m

Markschemeevidence of solving (M1)

e.g. equation, graph

(minutes) A2 N3

[3 marks]

Examiners reportIn part (d), candidates were clear on what was required and set their equation equal to 96. Yet again however, solving this equationgraphically using a GDC proved too daunting a task for most.

h(t) = 96

t = 12.8

32a. [3 marks]

There is a vertical tower TA of height 36 m at the base A of a hill. A straight path goes up the hill from A to a point U. This information isrepresented by the following diagram.

The path makes a angle with the horizontal.

The point U on the path is away from the base of the tower.

The top of the tower is fixed to U by a wire of length .

Complete the diagram, showing clearly all the information above.

Markscheme

A1A1A1 N3

Note: Award A1 for labelling with horizontal, A1 for labelling [AU] 25 metres, A1 for drawing [TU].

[3 marks]

4∘

25 m

x m

4∘

Examiners reportThis question was attempted in a satisfactory manner. Even the weakest candidates earned some marks here, showing some clearworking. In part (a) the diagram was completed fairly well, with some candidates incorrectly labelling the angle with the vertical as

. The cosine rule was applied satisfactory in part (b), although some candidates incorrectly used their calculators in radian mode.Approaches using a combination of the sine rule and/or right-angled triangle trigonometry were seen, especially when candidatesincorrectly labelled the path as being the distance from the horizontal to U.

4∘

25 m

32b. [4 marks]Find x .

Markscheme (A1)



e.g.

A1 N3

[4 marks]

Examiners reportThis question was attempted in a satisfactory manner. Even the weakest candidates earned some marks here, showing some clearworking. In part (a) the diagram was completed fairly well, with some candidates incorrectly labelling the angle with the vertical as

. The cosine rule was applied satisfactory in part (b), although some candidates incorrectly used their calculators in radian mode.Approaches using a combination of the sine rule and/or right-angled triangle trigonometry were seen, especially when candidatesincorrectly labelled the path as being the distance from the horizontal to U.

T U =A 86∘

= + − 2(25)(36)cosx2 252 362 86∘

x = 42.4

4∘

25 m

33a. [5 marks]

The diagram below shows a plan for a window in the shape of a trapezium.

Three sides of the window are long. The angle between the sloping sides of the window and the base is , where .

Show that the area of the window is given by .

2 m θ 0 < θ < π

2

y = 4 sin θ + 2 sin 2θ

Markschemeevidence of finding height, h (A1)

e.g. ,

evidence of finding base of triangle, b (A1)

e.g. ,

attempt to substitute valid values into a formula for the area of the window (M1)

e.g. two triangles plus rectangle, trapezium area formula

correct expression (must be in terms of ) A1

e.g. ,

attempt to replace by M1

e.g.

AG N0

[5 marks]

Examiners reportAs the final question of the paper, this question was understandably challenging for the majority of the candidates. Part (a) wasgenerally attempted, but often with a lack of method or correct reasoning. Many candidates had difficulty presenting their ideas in aclear and organized manner. Some tried a "working backwards" approach, earning no marks.

sin θ = h

22 sin θ

cosθ = b

22 cosθ

θ

2 ( × 2 cosθ × 2 sin θ) + 2 × 2 sin θ12

(2 sin θ)(2 + 2 + 4 cosθ)12

2 sin θ cosθ sin 2θ

4 sin θ + 2(2 sin θ cosθ)

y = 4 sin θ + 2 sin 2θ

33b. [4 marks]Zoe wants a window to have an area of . Find the two possible values of .

Markschemecorrect equation A1

e.g. ,

evidence of attempt to solve (M1)

e.g. a sketch,

, A1A1 N3

[4 marks]

Examiners reportIn part (b), most candidates understood what was required and set up an equation, but many did not make use of the GDC and insteadattempted to solve this equation algebraically which did not result in the correct solution. A common error was finding a secondsolution outside the domain.

5 m2 θ

y = 5 4 sin θ + 2 sin 2θ = 5

4 sin θ + 2 sin θ − 5 = 0

θ = 0.856 ( )49.0∘ θ = 1.25 ( )71.4∘

33c. [7 marks]John wants two windows which have the same area A but different values of .

Find all possible values for A .

θ

Markschemerecognition that lower area value occurs at (M1)

finding value of area at (M1)

e.g. , draw square

(A1)

recognition that maximum value of y is needed (M1)

(A1)

(accept ) A2 N5

[7 marks]

Examiners reportA pleasing number of stronger candidates made progress on part (c), recognizing the need for the end point of the domain and/or themaximum value of the area function (found graphically, analytically, or on occasion, geometrically). However, it was evident fromcandidate work and teacher comments that some candidates did not understand the wording of the question. This has been taken intoconsideration for future paper writing.

θ = π

2

θ = π

2

4 sin ( ) + 2 sin (2 × )π

2π

2

A = 4

A = 5.19615…

4 < A < 5.20 4 < A < 5.19

34. [7 marks]Solve , for .

Markschemeevidence of substituting for (M1)

evidence of substituting into (M1)

correct equation in terms of (seen anywhere) A1

e.g. ,

evidence of appropriate approach to solve (M1)

e.g. factorizing, quadratic formula

appropriate working A1

e.g. , ,

correct solutions to the equation

e.g. , , , (A1)

A1 N4

[7 marks]

Examiners reportThis question was quite difficult for most candidates. A number of students earned some credit for manipulating the equation withidentities, but many earned no further marks due to algebraic errors. Many did not substitute for ; others did this substitutionbut then did nothing further.

Few candidates were able to get a correct equation in terms of and many who did get the equation didn't know what to do withit. Candidates who correctly solved the resulting quadratic usually found the one correct value of , earning full marks.

cos2x − 3 cosx − 3 − x = xcos2 sin2 0 ≤ x ≤ 2π

cos2x

x + x = 1sin2 cos2

cosx

2 x − 1 − 3 cosx − 3 = 1cos2 2 x − 3 cosx − 5 = 0cos2

(2 cosx − 5)(cosx + 1) = 0 (2x − 5)(x + 1) cosx = 3± 49√4

cosx = 52

cosx = −1 x = 52

x = −1

x = π

cos2x

cosx

x

35a. [3 marks]

The following diagram shows a circle with centre O and radius 4 cm.

The points A, B and C lie on the circle. The point D is outside the circle, on (OC).

Angle ADC = 0.3 radians and angle AOC = 0.8 radians.

Find AD.

Markschemechoosing sine rule (M1)


e.g.

A1 N2

[3 marks]

Examiners reportThis question was generally quite well done, and it was pleasing to note that candidates could come up with multiple methods toarrive at the correct answers. Many candidates worked comfortably with the sine and cosine rules to find sides of triangles. Somecandidates chose alternative right-angled triangle methods, often with success, although this proved a time-consuming approach.Some unnecessarily converted the radian values to degrees, which sometimes led to calculation errors that could have been avoided.A large number of candidates accrued the accuracy penalty in this question.

=ADsin 0.8

4sin 0.3

AD = 9.71 (cm)

35b. [4 marks]Find OD.

MarkschemeMETHOD 1

finding angle (seen anywhere) (A1)

choosing cosine rule (M1)


e.g.

A1 N3

METHOD 2

finding angle (seen anywhere) (A1)

choosing sine rule (M1)


e.g.

A1 N3

[4 marks]


OAD = π − 1.1 = (2.04)

O = + − 2 × 9.71 × 4 × cos(π − 1.1)D2 9.712 42

OD = 12.1 (cm)

OAD = π − 1.1 = (2.04)

= =ODsin(π−1.1)

9.71sin 0.8

4sin 0.3

OD = 12.1 (cm)

35c. [2 marks]Find the area of sector OABC.

Markschemecorrect substitution into area of a sector formula (A1)

e.g.

A1 N2

[2 marks]


area = 0.5 × × 0.842

area = 6.4 (c )m2

35d. [4 marks]Find the area of region ABCD.

Markschemesubstitution into area of triangle formula OAD (M1)


e.g. , ,

subtracting area of sector OABC from area of triangle OAD (M1)

e.g.

A1 N2

[4 marks]


A= × 4 × 12.1 × sin 0.812

A= × 4 × 9.71 × sin 2.0412

A= × 12.1 × 9.71 × sin 0.312

area ABCD = 17.3067 − 6.4

area ABCD = 10.9 (c )m2

36. [4 marks]

Let and for .

Consider the graph of . Write down

(i) the x-intercept that lies between and ;

(ii) the period;

(iii) the amplitude.

Markscheme(i) (accept ) A1 N1

(ii) A2 N2

(iii) A1 N1

[4 marks]

Examiners reportMany candidates found parts (b) and (c) accessible, although quite a few did not know how to find the period of the cosine function.

f(x) = 5 cos xπ

4g(x) = −0.5 + 5x − 8x2 0 ≤ x ≤ 9

f

x = 0 x = 3

(2, 0) x = 2

period = 8

amplitude = 5

37a. [2 marks]

A rectangle is inscribed in a circle of radius 3 cm and centre O, as shown below.

The point P(x , y) is a vertex of the rectangle and also lies on the circle. The angle between (OP) and the x-axis is radians, where .

Write down an expression in terms of for

(i) ;

(ii) .

Markscheme(i) A1 N1

(ii) A1 N1

[2 marks]

Examiners reportCandidates familiar with the circular nature of sine and cosine found part (a) accessible. However, a good number of candidates leftthis part blank, which suggests that there was difficulty interpreting the meaning of the x and y in the diagram.

θ 0 ≤ θ ≤ π

2

θ

x

y

x = 3 cosθ

y = 3 sin θ

37b. [3 marks]Let the area of the rectangle be A.

Show that .

Markschemefinding area (M1)

e.g. ,

substituting A1

e.g. ,

A1

AG N0

[3 marks]

A = 18 sin 2θ

A = 2x × 2y A = 8 × bh12

A = 4 × 3 sin θ × 3 cosθ 8 × × 3 cosθ × 3 sin θ12

A = 18(2 sin θ cosθ)

A = 18 sin 2θ

Examiners reportThose with answers from (a) could begin part (b), but many worked backwards and thus earned no marks. In a "show that" question,a solution cannot begin with the answer given. The area of the rectangle could be found by using , or by using the eightsmall triangles, but it was essential that the substitution of the double-angle formula was shown before writing the given answer.

2x × 2y

38. [6 marks]

The vertices of the triangle PQR are defined by the position vectors

, and .

(i) Find .

(ii) Hence, find the area of triangle PQR, giving your answer in the form .

Markscheme(i) METHOD 1

evidence of appropriate approach (M1)

e.g. using , diagram

substituting correctly (A1)

e.g.

A1 N3

METHOD 2

since , (A1)

evidence of approach

e.g. drawing a right triangle, finding the missing side (A1)

A1 N3

(ii) evidence of appropriate approach (M1)

e.g. attempt to substitute into


e.g. area A1

area A1 N2

[6 marks]

Examiners reportMany candidates attained the value for sine in (c) with little difficulty, some using the Pythagorean identity, while others knew theside relationships in a 30-60-90 triangle. Unfortunately, a good number of candidates then used the side values of to find thearea of PQR , instead of the magnitudes of the vectors found in (a). Furthermore, the "hence" command was sometimes neglected asthe value of sine was expected to be used in the approach.

=OP−→ ⎛

⎝⎜4

−31

⎞⎠⎟ =OQ

−→− ⎛⎝⎜

3−12

⎞⎠⎟ =OR

−→− ⎛⎝⎜

6−15

⎞⎠⎟

sinR QP

a 3√

si R Q + co R Q = 1n2 P s2 P

sinR Q =P 1 − ( )12

2− −−−−−−√sinR Q =P 3

4

−−√ (= )3√2

cos =P 12

=P 60∘

sin =P 3√2

ab sin C12

= × ×12

6√ 24−−√ 3√2

= 3 3√

1,2, 3√

39a. [3 marks]

The circle shown has centre O and radius 3.9 cm.

Points A and B lie on the circle and angle AOB is 1.8 radians.

Find AB.

MarkschemeMETHOD 1

choosing cosine rule (M1)

substituting correctly A1

e.g.

(cm) A1 N2

METHOD 2

evidence of approach involving right-angled triangles (M1)


e.g. ,

(cm) A1 N2

METHOD 3

choosing the sine rule (M1)


e.g.

(cm) A1 N2

[3 marks]

Examiners reportThis question was well answered by the majority of candidates. Full solutions were common in both parts, and a variety of successfulapproaches were used. Radians were well handled with few candidates working with the angle in degrees. Some candidatesincorrectly found the length of the arc subtended by the central angle rather than the length of segment [AB].

AB = + − 2(3.9)(3.9)cos1.83.92 3.92− −−−−−−−−−−−−−−−−−−−−−−√

AB = 6.11

sin 0.9 = x

3.9AB = 3.9sin 0.91

2

AB = 6.11

=sin 0.670…3.9

sin 1.8AB

AB = 6.11


MarkschemeMETHOD 1

reflex (A2)


area =34.1 (cm ) A1 N2

METHOD 2

finding area of circle (A1)

finding area of (minor) sector (A1)

subtracting M1

e.g. ,

area = 34.1 (cm ) A1 N2

METHOD 3

finding reflex (A2)

finding proportion of total area of circle A1

e.g. ,

area = 34.1 (cm ) A1 N2

[4 marks]

Examiners reportIn part (b), some candidates incorrectly subtracted the area of the triangle or even a length. Many candidates failed to give answers to3 significant figures and therefore lost an accuracy mark.

A B = 2π − 1.8O (= 4.4832)

A = (3.9 (4.4832…)12

)2

2

A = π(3.9)2 (= 47.78…)

A = (3.9 (1.8)12

)2 (= 13.68…)

π(3.9 − 0.5(3.9 (1.8))2 )2 47.8 − 13.7

2

A B = 2π − 1.8O (= 4.4832)

× π(3.92π−1.82π

)2 × πθ

2πr2

2

40a. [1 mark]

Let , . Let .

Write down the period of .

Markschemeperiod is A1 N1

[1 mark]

Examiners reportThe majority of candidates handled the composition of the two given functions well. However, a large number of candidates haddifficulties simplifying the result correctly. The period and range of the resulting trig function was not handled well. If candidatesknew the definition of "range", they often did not express it correctly. Many candidates correctly used their GDCs to find the periodand range, but this approach was not the most efficient.

f(x) = + 13x

2g(x) = 4 cos( ) − 1x

3h(x) = (g ∘ f)(x)

h

4π(12.6)

40b. [2 marks]Write down the range of .

Markschemerange is A1A1 N2

[2 marks]

h

−5 ≤ h(x) ≤ 3 ([−5,3])

Examiners reportThe majority of candidates handled the composition of the two given functions well. However, a large number of candidates haddifficulties simplifying the result correctly. The period and range of the resulting trig function was not handled well. If candidatesknew the definition of "range", they often did not express it correctly. Many candidates correctly used their GDCs to find the periodand range, but this approach was not the most efficient.

41. [6 marks]Let , for . Solve the equation .

Markscheme (A1)

not possible (seen anywhere) (A1)

simplifying

e.g. , , A1

EITHER

A1

A2 N4

OR

sketch of , , triangle with sides , , A1

work leading to A1

verifying satisfies equation A1 N4

[6 marks]

Examiners reportThose who realized was a common factor usually earned the first four marks. Few could reason with the given information tosolve the equation from there. There were many candidates who attempted some fruitless algebra that did not include factorization.

f(x) = sin x + cosx3√ e2x e2x 0 ≤ x ≤ π f(x) = 0

( sin x + cosx) = 0e2x 3√

= 0e2x

sin x + cosx = 03√ sin x = −cosx3√ =sin x

− cos x13√

tanx = − 13√

x = 5π

6

30∘ 60∘ 90∘ 1 2 3√

x = 5π

6

5π

6

e2x

42. [4 marks]

Let and .

Let . Find the exact value of .

Markschemeevidence of choosing product rule (M1)

e.g.

correct expression A1

e.g.

complete correct substitution of (A1)

e.g. � � � � � � � � �

A1 N3

[4 marks]

f(x) = e−3x g(x) = sin (x − )π

3

h(x) = sin (x − )e−3x π

3( )h′ π

3

u + vv′ u′

−3 sin (x − ) + cos(x − )e−3x π

3e−3x π

3

x = π

3

−3 sin ( − ) + cos( − )e−3 π

3π

3π

3e−3 π

3π

3π

3

( ) =h′ π

3e−π

Examiners reportOften the substitution of was incomplete or not done at all.π

3

43a. [3 marks]

The diagram below shows a triangle ABD with AB =13 cm and AD = 6.5 cm.

Let C be a point on the line BD such that BC = AC = 7 cm.

Find the size of angle ACB.

MarkschemeMETHOD 1

evidence of choosing the cosine formula (M1)


e.g.

radians A1 N2

METHOD 2

evidence of appropriate approach involving right-angled triangles (M1)


e.g.

radians A1 N2

[3 marks]

Examiners reportThis question was generally well done. Even the weakest candidates often earned marks. Only a very few candidates used a right-angled triangle approach.

cosA B =C + −72 72 132

2×7×7

A B = 2.38C (= )136∘

sin ( A B) =12

C 6.57

A B = 2.38C (= )136∘

43b. [5 marks]Find the size of angle CAD.

MarkschemeMETHOD 1

(A1)

evidence of choosing the sine rule in triangle ACD (M1)


e.g.

A1

A1 N3

METHOD 2

(A1)

evidence of choosing the sine rule in triangle ABD (M1)


e.g.

A1

A1 N3

Note: Two triangles are possible with the given information. If candidate finds leading to , award marks as per markscheme.

[5 marks]

Examiners reportAlmost no candidates realized there was an ambiguous case of the sine rule in part (b). Those who did not lose the mark for accuracyin the previous question often lost it here.

A D = π − 2.381C (180 − 136.4)

=6.5sin 0.760…

7

sin A CD

A C = 0.836…D (= 47.9 )…∘

C D = π − (0.760… + 0.836…)A (180 − (43.5… + 47.9…))

= 1.54 (= )88.5∘

A C = (π − 2.381)B 12

( (180 − 136.4))12

=6.5sin 0.380…

13

sin A CD

A C = 0.836…D (= 47.9 )…∘

C D = π − 0.836… − (π − 2.381…)A (= 180 − 47.9… − (180 − 136.4))

= 1.54 (= )88.5∘

A C = 2.31D ( )132∘ C D = 0.076A( )4.35∘

44a. [3 marks]

Let , for .

Hence write in the form .

Markscheme (accept , , ) A1A1A1 N3

[3 marks]

Examiners reportA number of candidates had difficulty writing the correct value of q in part (c).

f(x) = 3 sin x + 4 cosx −2π ≤ x ≤ 2π

f(x) p sin(qx + r)

f(x) = 5 sin(x + 0.927) p = 5 q = 1 r = 0.927

44b. [2 marks]Write down the two values of k for which the equation has exactly two solutions.


[2 marks]

f(x) = k

k = −5 k = 5

Examiners reportPart (e) proved challenging for many candidates, although if candidates answered this part, they generally did so correctly.

45a. [6 marks]

The graph of , for , is shown below.

There is a minimum point at (0, −3) and a maximum point at (4, 7) .

Find the value of

(i) p ;

(ii) q ;

(iii) r.

Markscheme(i) evidence of finding the amplitude (M1)

e.g. , amplitude

A1 N2

(ii) period (A1)

A1 N2

(iii) (A1)

A1 N2

[6 marks]

Examiners reportMany candidates did not recognize that the value of p was negative. The value of q was often interpreted incorrectly as the period butmost candidates could find the value of r, the vertical translation.

y = p cosqx + r −5 ≤ x ≤ 14

7+32

= 5

p = −5

= 8

q = 0.785 (= = )2π

8π

4

r = 7−32

r = 2

45b. [1 mark]The equation has exactly two solutions. Write down the value of k.

Markscheme (accept ) A1 N1

[1 mark]

y = k

k = −3 y = −3

Examiners reportIn part (b), candidates either could not find a solution or found too many.

46a. [1 mark]

The diagram below shows a quadrilateral ABCD with obtuse angles and .

AB = 5 cm, BC = 4 cm, CD = 4 cm, AD = 4 cm , , , .

Use the cosine rule to show that .

Markschemecorrect substitution A1

e.g. ,

AG

[1 mark]

Examiners reportMany candidates worked comfortably with the sine and cosine rules in part (a) and (b).

A CB A CD

B C =A 30∘ A C =B x∘ A C =D y∘

AC = 41 − 40 cosx− −−−−−−−−−√

25 + 16 − 40 cosx + − 2 × 4 × 5 cosx52 42

AC = 41 − 40 cosx− −−−−−−−−−√

46b. [2 marks]Use the sine rule in triangle ABC to find another expression for AC.

Markschemecorrect substitution A1

e.g. ,

(accept ) A1 N1

[2 marks]

Examiners reportMany candidates worked comfortably with the sine and cosine rules in part (a) and (b). Equally as many did not take the cue from theword "hence" and used an alternate method to solve the problem and thus did not receive full marks. Those who managed to set upan equation, again did not go directly to their GDC but rather engaged in a long, laborious analytical approach that was usuallyunsuccessful.

=ACsin x

4sin 30

AC = 4 sin x12

AC = 8 sin x 4 sin x

sin 30

46c. [6 marks](i) Hence, find x, giving your answer to two decimal places.

(ii) Find AC .

Markscheme(i) evidence of appropriate approach using AC M1

e.g. , sketch showing intersection

correct solution , (A1)

obtuse value (A1)

to 2 dp (do not accept the radian answer 1.94 ) A1 N2

(ii) substituting value of x into either expression for AC (M1)

e.g.

A1 N2

[6 marks]

Examiners reportEqually as many did not take the cue from the word "hence" and used an alternate method to solve the problem and thus did notreceive full marks. Those who managed to set up an equation, again did not go directly to their GDC but rather engaged in a long,laborious analytical approach that was usually unsuccessful. No matter what values were found in (c) (i) most candidates recoveredand earned follow through marks for the remainder of the question. A large number of candidates worked in the wrong mode androunded prematurely throughout this question often resulting in accuracy penalties.

8 sin x = 41 − 40 cosx− −−−−−−−−−√

8.682… 111.317…

111.317…

x = 111.32

AC = 8 sin 111.32

AC = 7.45

46d. [5 marks](i) Find y.

(ii) Hence, or otherwise, find the area of triangle ACD.

Markscheme(i) evidence of choosing cosine rule (M1)

e.g.


e.g. , ,

A1 N2

(ii) correct substitution into area formula (A1)

e.g. ,

area A1 N2

[5 marks]

Examiners reportEqually as many did not take the cue from the word "hence" and used an alternate method to solve the problem and thus did notreceive full marks. Those who managed to set up an equation, again did not go directly to their GDC but rather engaged in a long,laborious analytical approach that was usually unsuccessful. No matter what values were found in (c) (i) most candidates recoveredand earned follow through marks for the remainder of the question. A large number of candidates worked in the wrong mode androunded prematurely throughout this question often resulting in accuracy penalties.

cosB = + −a2 c2 b2

2ac

+ −42 42 7.452

2×4×4= 32 − 32 cosy7.452 cosy = −0.734…

y = 137

× 4 × 4 × sin 13712

8 sin 137

= 5.42

47a. [2 marks]

The diagram below shows a circle with centre O and radius 8 cm.

The points A, B, C, D, E and F are on the circle, and [AF] is a diameter. The length of arc ABC is 6 cm.

Find the size of angle AOC .

Markschemeappropriate approach (M1)

e.g.

A1 N2

[2 marks]

Examiners reportMost candidates demonstrated understanding of trigonometry on this question. They generally did well in parts (a) and (c), and evenmany of them on part (b). Fewer candidates could do part (d).

Many opted to work in degrees rather than in radians, which often introduced multiple inaccuracies. Some worked with an incorrectradius of 6 or 10.

A pleasing number knew how to find the area of the shaded region.

Inability to work in radians and misunderstanding of significant figures were common problems, though. Weaker candidates oftenmade the mistake of using triangle formulae for sectors or used degrees in the formulas instead of radians.

For some candidates there were many instances of confusion between lines and arcs. In (a) some treated 6 as the length of AC . In (d)some found the length of arc EF rather than the length of the segment.

Several students seemed to confuse the area of sector in (b) with the shaded region.

6 = 8θ

A C = 0.75O

47b. [6 marks]Hence find the area of the shaded region.

Markschemeevidence of substitution into formula for area of triangle (M1)

e.g.

area (A1)

evidence of substitution into formula for area of sector (M1)

e.g.

area of sector (A1)

evidence of substituting areas (M1)

e.g. ,

area of shaded region A1 N4

[6 marks]







area = × 8 × 8 × sin(0.75)12

= 21.8…

area = × 64 × 0.7512

= 24

θ − ab sin C12

r2 12

area of sector − area of triangle

= 2.19 cm2

47c. [2 marks]The area of sector OCDE is .

Find the size of angle COE .

Markschemeattempt to set up an equation for area of sector (M1)

e.g.

(1.41 to 3 sf) A1 N2

[2 marks]







45 cm2

45 = × × θ12

82

C E = 1.40625O

47d. [5 marks]Find EF .

MarkschemeMETHOD 1

attempting to find angle EOF (M1)

e.g.

(seen anywhere) A1



e.g.

EF A1 N3

METHOD 2

attempting to find angles that are needed (M1)

e.g. angle EOF and angle OEF

and A1



e.g.

EF A1 N3

METHOD 3

attempting to find angle EOF (M1)

e.g.

(seen anywhere) A1

evidence of using half of triangle EOF (M1)

e.g.

correct calculation A1

e.g.

EF A1 N3

[5 marks]







π − 0.75 − 1.41

E F = 0.985O

EF = + − 2 × 8 × 8 × cos0.98582 82− −−−−−−−−−−−−−−−−−−−−−−√

= 7.57 cm

E F = 0.9853…O O F (or O E) = 1.078…E F

=EFsin 0.985

8sin 1.08

= 7.57 cm

π − 0.75 − 1.41

E F = 0.985O

x = 8 sin 0.9852

x = 3.78

= 7.57 cm

48a. [2 marks]

Let .

Show that .

f(x) = x + x tanx, < x < πsin3 cos3 π

2

f(x) = sin x

Markschemechanging into A1

e.g.

simplifying A1

e.g ,

AG N0

[2 marks]

Examiners reportNot surprisingly, this question provided the greatest challenge in section A. In part (a), candidates were able to use the identity

, but many could not proceed any further.

tanx sin x

cos x

x + xsin3 cos3 sin x

cos x

sin x( x + x)sin2 cos2 x + sin x − xsin3 sin3

f(x) = sin x

tanx = sin x

cos x

48b. [5 marks]Let . Show that .

Markschemerecognizing , seen anywhere (A1)

evidence of using double angle identity , seen anywhere (M1)

evidence of using Pythagoras with M1

e.g. sketch of right triangle,

(accept ) (A1)

A1

AG N0

[5 marks]

Examiners reportPart (b) was generally well done by those candidates who attempted it, the major error arising when the negative sign "magically"appeared in the answer. Many candidates could find the value of cosx but failed to observe that cosine is negative in the givendomain.

sin x = 23

f(2x) = − 4 5√9

f(2x) = sin 2x

sin(2x) = 2 sin xcosx

sin x = 23

x + x = 1sin2 cos2

cosx = − 5√3

5√3

f(2x) = 2 ( ) (− )23

5√3

f(2x) = − 4 5√9

f(t) = acosb(t − c) + d y = f(t)

49a. [7 marks]

Let , . Part of the graph of is given below.

When , there is a maximum value of 29, at M.

When , there is a minimum value of 15.

(i) Find the value of a.

(ii) Show that .

(iii) Find the value of d.

(iv) Write down a value for c.

Markscheme(i) attempt to substitute (M1)

e.g.

(accept ) A1 N2

(ii) (A1)

A1

AG N0

(iii) attempt to substitute (M1)

e.g.

A1 N2

(iv) (accept from ) A1 N1

Note: Other correct values for c can be found, , .

[7 marks]

Examiners reportThis question was the most difficult on the paper. Where candidates attempted this question, part (a) was answered satisfactorily.

f(t) = acosb(t − c) + d t ≥ 0 y = f(t)

t = 3

t = 9

b = π

6

a = 29−152

a = 7 a = −7

period = 12

b = 2π

12

b = π

6

d = 29+152

d = 22

c = 3 c = 9 a = −7

c = 3 ± 12k k ∈ Z

49b. [2 marks]The transformation P is given by a horizontal stretch of a scale factor of , followed by a translation of .

Let be the image of M under P. Find the coordinates of .

12

( )3−10

M ′ M ′

Markschemestretch takes 3 to 1.5 (A1)

translation maps to (so is ) A1 N2

[2 marks]

Examiners reportFew answered part (b) correctly as most could not interpret the horizontal stretch.

(1.5, 29) (4.5, 19) M ′ (4.5, 19)

49c. [4 marks]The graph of g is the image of the graph of f under P.

Find in the form .

Markscheme A1A2A1 N4

Note: Award A1 for , A2 for 4.5, A1 for 12.

Other correct values for c can be found, , .

[4 marks]

Examiners reportFew answered part (b) correctly as most could not interpret the horizontal stretch. As a result, there were many who were unable toanswer part (c) although follow through marks were often obtained from incorrect answers in both parts (a) and (b). The link betweenthe answer in (b) and the value of C in part (c) was lost on all but the most attentive.

g(t) g(t) = 7 cosB(t − c) + D

g(t) = 7 cos (t − 4.5) + 12π

3

π

3

c = 4.5 ± 6k k ∈ Z

49d. [3 marks]The graph of g is the image of the graph of f under P.

Give a full geometric description of the transformation that maps the graph of g to the graph of f .

Markschemetranslation (A1)

horizontal stretch of a scale factor of 2 (A1)

completely correct description, in correct order A1 N3

e.g. translation then horizontal stretch of a scale factor of 2

[3 marks]

Examiners reportIn part (d), some candidates could name the transformations required, although only a handful provided the correct order of thetransformations to return the graph to its original state.

( )−310

( )−310

50a. [4 marks]

A ship leaves port A on a bearing of . It sails a distance of to point B. At B, the ship changes direction to a bearing of . Itsails a distance of to reach point C. This information is shown in the diagram below.

A second ship leaves port A and sails directly to C.

Find the distance the second ship will travel.

Markschemefinding ( radians) (A1)


e.g.


e.g.

(km) A1

Examiners reportA good number of candidates found this question very accessible, although some attempted to use Pythagoras' theorem to find AC.

030∘ 25 km 100∘

40 km

A C =B 110∘ = 1.92

A = A + B − 2(AB)(BC)cosA CC2 B2 C2 B

A = + − 2(25)(40)cosC2 252 402 110∘

A = 53.9C

50b. [3 marks]Find the bearing of the course taken by the second ship.

MarkschemeMETHOD 1

correct substitution into the sine rule A1

e.g. A1

bearing A1 N1

METHOD 2

correct substitution into the cosine rule A1

e.g. A1

bearing A1 N1

[3 marks]

=sin B CA40

sin 110∘

53.9

B C =A 44.2∘

= 074∘

cosB C =A − −402 252 53.92

−2(25)(53.9)

B C =A 44.3∘

= 074∘

Examiners reportOften candidates correctly found in part (b), but few added the to obtain the required bearing. Some candidates calculated

, misinterpreting that the question required the course of the second ship.B CA 30∘

B AC

51. [3 marks]

The diagram shows a parallelogram ABCD.

The coordinates of A, B and D are A(1, 2, 3) , B(6, 4,4 ) and D(2, 5, 5) .

Hence, or otherwise, find the area of the parallelogram.

MarkschemeMETHOD 1

evidence of using (M1)


e.g.

A1 N2

METHOD 2


finding height of parallelogram A1

e.g. ,

A1 N2

[3 marks]

Examiners reportFew candidates considered that the area of the parallelogram is twice the area of a triangle, which is conveniently found using .In an effort to find base height , many candidates multiplied the magnitudes of and , missing that the height of aparallelogram is perpendicular to a base.

area = 2 ( sin D B)12

∣∣AD−→− ∣

∣∣∣AB−→∣

∣ A

area = 2 ( (3,741 …)(5.477…)sin 0.883…)12

area = 15.8

area = b × h

h = 3.741… × sin 0.883… (= 2.892…) h = 5.477… × sin 0.883… (= 4.234…)

area = 15.8

B DA

× AB−→

AD−→−

52a. [2 marks]

Let and . Give your answers to the following in terms of p and/or q .

Write down an expression for

(i) ;

(ii) .

p = sin 40∘ q = cos110∘

sin 140∘

cos70∘

Markscheme(i) A1 N1

(ii) A1 N1

[2 marks]

Examiners reportThis was one of the most difficult problems for the candidates. Even the strongest candidates had a hard time with this one and only afew received any marks at all.

sin = p140∘

cos = −q70∘

52b. [3 marks]Find an expression for .

MarkschemeMETHOD 1


e.g. diagram, (seen anywhere)

(A1)

A1 N2

METHOD 2


(A1)

A1 N2

[3 marks]

Examiners reportMany did not appear to know the relationships between trigonometric functions of supplementary angles and that the use of

results in a value. The application of a double angle formula also seemed weak.

cos140∘

θ + θ = 1sin2 cos2

1 − p2− −−−−√

cos = ±140∘ 1 − p2− −−−−√

cos = −140∘ 1 − p2− −−−−√

cos2θ = 2 θ − 1cos2

cos = 2 70 − 1140∘ cos2

cos = 2(−q − 1140∘ )2 (= 2 − 1)q2

x + x = 1sin2 cos2 ±

52c. [1 mark]Find an expression for .

MarkschemeMETHOD 1

A1 N1

METHOD 2

A1 N1

[1 mark]

Examiners reportThis was one of the most difficult problems for the candidates. Even the strongest candidates had a hard time with this one and only afew received any marks at all. Many did not appear to know the relationships between trigonometric functions of supplementaryangles and that the use of results in a value. The application of a double angle formula also seemed weak.

tan140∘

tan = = −140∘ sin 140∘

cos 140∘

p

1−p2√

tan =140∘ p

2 −1q2

x + x = 1sin2 cos2 ±

g(x) = 3 sin 2x

53a. [1 mark]

Consider .

Write down the period of g.

Markscheme A1 N1

[1 mark]

Examiners reportMany candidates were unable to write down the period of the function.

g(x) = 3 sin 2x

period = π

53b. [3 marks]On the diagram below, sketch the curve of g, for .

Markscheme

A1A1A1 N3

Note: Award A1 for amplitude of 3, A1 for their period, A1 for a sine curve passing through and .

[3 marks]

Examiners reportMany candidates were unable to write down the period of the function. However, they were often then able to go and correctlysketch the graph with the correct period.

0 ≤ x ≤ 2π

(0, 0) (0, 2π)

53c. [2 marks]Write down the number of solutions to the equation , for .g(x) = 2 0 ≤ x ≤ 2π

Markschemeevidence of appropriate approach (M1)

e.g. line on graph, discussion of number of solutions in the domain

4 (solutions) A1 N2

[2 marks]

Examiners reportThe final part was poorly done with many candidates finding the number of zeros instead of the intersection with the line .

y = 2

y = 2

54a. [3 marks]

The diagram below shows triangle PQR. The length of [PQ] is 7 cm , the length of [PR] is 10 cm , and is .

Find .

Markschemechoosing sine rule (M1)


A1 N2

[3 marks]

Examiners reportThis question was well done with most students using the law of sines to find the angle.

P RQ 75∘

P QR

=sin R

7sin 75∘

10

sin R = 0.676148…

P Q = 42R .5∘

54b. [3 marks]Find the area of triangle PQR.

Markscheme

(A1)

substitution into any correct formula A1

e.g.

(cm ) A1 N2

[3 marks]

P = 180 − 75 − R

P = 62.5

area ΔPQR = × 7 × 10 × sin(their P)12

= 31.0 2

Examiners reportIn part (b), the most common error occurred when angle R or 75 degrees was used to find the area. This particular question was themost common place to incur an accuracy penalty.

55a. [2 marks]

The diagram below shows a circle centre O, with radius r. The length of arc ABC is and .

Find the value of r.

Markschemeevidence of appropriate approach M1

e.g.

(cm) A1 N1

[2 marks]

Examiners reportFew errors were made in this question. Those that were made were usually arithmetical in nature.

3π cm A C =O 2π

9

3π = r 2π

9

r = 13.5

55b. [2 marks]Find the perimeter of sector OABC.

Markschemeadding two radii plus (M1)

(cm) ( ) A1 N2

[2 marks]


3π

perimeter = 27 + 3π = 36.4

55c. [2 marks]Find the area of sector OABC.

Markschemeevidence of appropriate approach M1

e.g.

area ( ) ( ) A1 N1

[2 marks]


× ×12

13.52 2π

9

= 20.25π cm2 = 63.6

56a. [3 marks]Given that and , find .

Markschemeevidence of choosing the formula (M1)

Note: If they choose another correct formula, do not award the M1 unless there is evidence of finding


e.g. ,

A1 N2

[3 marks]

Examiners reportThis question was very poorly done, and knowledge of basic trigonometric identities and values of trigonometric functions of obtuseangles seemed distinctly lacking. Candidates who recognized the need of an identity for finding given seldom chose themost appropriate of the three and even when they did often used it incorrectly with expressions such as .

cosA = 13

0 ≤ A ≤ π

2cos2A

cos2A = 2 A − 1cos2

A = 1 −sin2 19

cos2A = −( )13

289

cos2A = 2 × − 1( )13

2

cos2A = − 79

cos2A cosA

2 − 1cos2 19

56b. [3 marks]Given that and , find .sin B = 23

≤ B ≤ ππ

2cosB

MarkschemeMETHOD 1


e.g. , (seen anywhere),

(A1)

A1 N2

METHOD 2

diagram M1

for finding third side equals (A1)

A1 N2

[3 marks]

Examiners reportThis question was very poorly done, and knowledge of basic trigonometric identities and values of trigonometric functions of obtuseangles seemed distinctly lacking. Candidates who recognized the need of an identity for finding given seldom chose themost appropriate of the three and even when they did often used it incorrectly with expressions such as .

B + B = 1sin2 cos2

+ B = 1( )23

2cos2 5

9

−−√cosB = ± 5

9

−−√ (= ± )5√3

cosB = − 59

−−√ (= − )5√3

5√

cosB = − 5√3

cos2A cosA

2 − 1cos2 19

57. [5 marks]

Let .

(i) Write down the range of the function f .

(ii) Consider , . Write down the number of solutions to this equation. Justify your answer.

Markscheme(i) range of f is , A2 N2

(ii) A1

justification for one solution on R1

e.g. , unit circle, sketch of

1 solution (seen anywhere) A1 N1

[5 marks]

Examiners reportThis question was not done well by most candidates. No more than one-third of them could correctly give the range of and few could provide adequate justification for there being exactly one solution to in the interval .

f : x ↦ xsin3

f(x) = 1 0 ≤ x ≤ 2π

[−1, 1] (−1 ≤ f(x) ≤ 1)

x ⇒ 1 ⇒ sin x = 1sin3

[0, 2π]

x = π

2sin x

f(x) = xsin3

f(x) = 1 [0, 2π]

58a. [2 marks]

The following diagram shows a semicircle centre O, diameter [AB], with radius 2.

Let P be a point on the circumference, with radians.

Find the area of the triangle OPB, in terms of .

Markschemeevidence of using area of a triangle (M1)

e.g.

A1 N2

[2 marks]

Examiners reportMost candidates could obtain the area of triangle OPB as equal to , though was given quite often as the area.

P B = θO

θ

A = × 2 × 2 × sin θ12

A = 2 sin θ

2 sin θ 2θ

58b. [3 marks]Explain why the area of triangle OPA is the same as the area triangle OPB.

MarkschemeMETHOD 1

(A1)

A1

since R1

then both triangles have the same area AG N0

METHOD 2

triangle OPA has the same height and the same base as triangle OPB R3

then both triangles have the same area AG N0

[3 marks]

Examiners reportA minority recognized the equality of the sines of supplementary angles and the term complementary was frequently used instead ofsupplementary. Only a handful of candidates used the simple equal base and altitude argument.

P A =π − θO

area ΔOPA = 2 × 2 × sin(π − θ)12

(= 2 sin(π − θ))

sin(π − θ) = sin θ

59a. [3 marks]

The following graph shows the depth of water, y metres , at a point P, during one day. The time t is given in hours, from midnight to noon.

Use the graph to write down an estimate of the value of t when

(i) the depth of water is minimum;

(ii) the depth of water is maximum;

(iii) the depth of the water is increasing most rapidly.

Markscheme(i) 7 A1 N1

(ii) 1 A1 N1

(iii) 10 A1 N1

[3 marks]

Examiners reportFor part (a), most candidates correctly used the graph to identify the times of maximum and minimum depth. Most failed to considerthat the depth of water is increasing most rapidly at a point of inflexion and often answered with the interval to . A fewcandidates answered with the depth instead of time, misinterpreting which axis to consider.

t = 9 t = 11

59b. [6 marks]The depth of water can be modelled by the function .

(i) Show that .

(ii) Write down the value of C.

(iii) Find the value of B.

y = cosA(B(t − 1)) + C

A = 8

Markscheme(i) evidence of appropriate approach M1

e.g.

AG N0

(ii) A2 N2

(iii) METHOD 1

(A1)

evidence of using (accept ) (M1)

e.g.

(accept 0.524 or 30) A1 N3

METHOD 2

evidence of substituting (M1)

e.g.

simplifying (A1)

e.g.

(accept 0.524 or 30) A1 N3

[6 marks]

Examiners reportA substantial number of candidates showed difficulty finding parameters of a trigonometric function with many only makingsuperficial attempts at part (b), often leaving it blank entirely.

Some divided by the period of 12, while others substituted an ordered pair such as and solved for B, often correctly. Manyfound that , thus confusing the vertical translation with a y-intercept.

A = 18−22

A = 8

C = 10

period = 12

B × period = 2π 360∘

12 = 2π

B

B = π

6

10 = 8 cos3B + 10

cos3B = 0 (3B = )π

2

B = π

6

2π (4, 10)c = 17

59c. [2 marks]A sailor knows that he cannot sail past P when the depth of the water is less than 12 m . Calculate the values of t betweenwhich he cannot sail past P.

Markschemecorrect answers A1A1

e.g. , , between 03:31 and 10:29 (accept 10:30) N2

[2 marks]

Examiners reportFor (c), many candidates simply read approximate values from the graph where and thus answered with and . Although the latter value is correct to three significant figures, incurs the accuracy penalty as it was expected that candidatescalculate this value in their GDC to achieve a result of . Those who attempted an analytic approach rarely achieved correctresults.

t = 3.52 t = 10.5

y = 12 t = 3.5 t = 10.5t = 3.5

t = 3.52

60a. [3 marks]

The expression can be expressed in the form .

Find the value of a and of b .

6 sin xcosx asin bx

Markschemerecognizing double angle M1

e.g. ,

, A1A1 N3

[3 marks]

Examiners report[N/A]

3 × 2 sin xcosx 3 sin 2x

a = 3 b = 2

60b. [4 marks]Hence or otherwise, solve the equation , for .

Markschemesubstitution M1

A1

finding the angle A1

e.g. ,

A1 N2

Note: Award A0 if other values are included.

[4 marks]


6 sin xcosx = 32

≤ x ≤π

4π

2

3 sin 2x = 32

sin 2x = 12

π

62x = 5π

6

x = 5π

12

61a. [3 marks]

Let , . The following diagram shows the graph of .

The -intercept is at ( , ) , there is a minimum point at A ( , ) and a maximum point at B.

Find the maximum value of .

f(x) = cosx + sin x3√ 0 ≤ x ≤ 2π f

y 0 1 p q

f(x)

Markschememax when R1

correctly substituting into A1

e.g.

max value is 2 A1 N1

[3 marks]


x = π

3

x = π

3f(x)

+ ( )12

3√ 3√2

61b. [2 marks]The function can be written in the form .

Write down the value of r and of a .


[2 marks]


f(x) rcos(x − a)

r = 2 a = π

3

62. [7 marks]

The following diagram shows a pole BT 1.6 m tall on the roof of a vertical building.

The angle of depression from T to a point A on the horizontal ground is .

The angle of elevation of the top of the building from A is .

Find the height of the building.

35∘

30∘

MarkschemeMETHOD 1

appropriate approach M1

e.g. completed diagram

attempt at set up A1

e.g. correct placement of one angle

, A1A1

attempt to set up equation M1

e.g. isolate x

correct equation A1

e.g.

A1 N3

METHOD 2

A1

in triangle ATB, , A1A1

choosing sine rule M1


e.g. A1

A1

A1 N3

[7 marks]


tan30 = h

xtan35 = h+1.6

x

=h

tan 30h+1.6tan 35

h = 7.52

sin 30 = h

l

=A 5∘ =T 55∘

=h/ sin 30sin 55

1.6sin 5

h = 1.6×sin 30×sin 55sin 5

h = 7.52

63a. [3 marks]

A circle centre O and radius is shown below. The chord [AB] divides the area of the circle into two parts. Angle AOB is .

Find an expression for the area of the shaded region.

r θ

Printed for sek-catalunya

© International Baccalaureate Organization 2014 International Baccalaureate® - Baccalauréat International® - Bachillerato Internacional®

Markschemesubstitution into formula for area of triangle A1

e.g.

evidence of subtraction M1

correct expression A1 N2

e.g. ,

[3 marks]


r × rsin θ12

θ − sin θ12

r2 12

r2 (θ − sin θ)12

r2

63b. [5 marks]The chord [AB] divides the area of the circle in the ratio 1:7. Find the value of .

Markschemeevidence of recognizing that shaded area is of area of circle M1

e.g. seen anywhere

setting up correct equation A1

e.g.

eliminating 1 variable M1

e.g. ,

attempt to solve M1

e.g. a sketch, writing

(do not accept degrees) A1 N1

[5 marks]


θ

18

18

(θ − sin θ) = π12

r2 18

r2

(θ − sin θ) = π12

18

θ − sin θ = π

4

sin x − x + = 0π

4

θ = 1.77

Documents

New test - October 20, 2014 [531 marks]